CodeForces Round #550 Div.3

http://codeforces.com/contest/1144

A. Diverse Strings

A string is called diverse if it contains consecutive (adjacent) letters of the Latin alphabet and each letter occurs exactly once. For example, the following strings are diverse: "fced", "xyz", "r" and "dabcef". The following string are not diverse: "az", "aa", "bad" and "babc". Note that the letters 'a' and 'z' are not adjacent.

Formally, consider positions of all letters in the string in the alphabet. These positions should form contiguous segment, i.e. they should come one by one without any gaps. And all letters in the string should be distinct (duplicates are not allowed).

You are given a sequence of strings. For each string, if it is diverse, print "Yes". Otherwise, print "No".

Input

The first line contains integer nn (1≤n≤1001≤n≤100), denoting the number of strings to process. The following nn lines contains strings, one string per line. Each string contains only lowercase Latin letters, its length is between 11 and 100100, inclusive.

Output

Print nn lines, one line per a string in the input. The line should contain "Yes" if the corresponding string is diverse and "No" if the corresponding string is not diverse. You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.

Example

input

Copy

8
fced
xyz
r
dabcef
az
aa
bad
babc

output

Copy

Yes
Yes
Yes
Yes
No
No
No
No

代码：

#include <bits/stdc++.h>
using namespace std;

int N;
map<char, int> mp;

int main() {
    scanf("%d", &N);
    getchar();
    for(int i = ; i < N; i ++) {
        mp.clear();
        string s;
        cin >> s;
        bool flag = true;
        int len = s.length();
        int minn = ;
        for(int j = ; j < len; j ++) {
            mp[s[j]] ++;
            minn = min(minn, s[j] - 'a');
            if(mp[s[j]] > ) flag = false;
        }
        //printf("!!!%d\n", minn);
        for(int j = minn; j < minn + len; j ++) {
            if(mp[j + 'a'] == ) {
                flag = false;
                break;
            }
        }
        if(flag) printf("Yes\n");
        else printf("No\n");
    }
    return ;
}

B. Parity Alternated Deletions

Polycarp has an array aa consisting of nn integers.

He wants to play a game with this array. The game consists of several moves. On the first move he chooses any element and deletes it (after the first move the array contains n−1n−1 elements). For each of the next moves he chooses any element with the only restriction: its parity should differ from the parity of the element deleted on the previous move. In other words, he alternates parities (even-odd-even-odd-... or odd-even-odd-even-...) of the removed elements. Polycarp stops if he can't make a move.

Formally:

• If it is the first move, he chooses any element and deletes it;
• If it is the second or any next move:
  • if the last deleted element was odd, Polycarp chooses any even element and deletes it;
  • if the last deleted element was even, Polycarp chooses any odd element and deletes it.
• If after some move Polycarp cannot make a move, the game ends.

Polycarp's goal is to minimize the sum of non-deleted elements of the array after end of the game. If Polycarp can delete the whole array, then the sum of non-deleted elements is zero.

Help Polycarp find this value.

Input

The first line of the input contains one integer nn (1≤n≤20001≤n≤2000) — the number of elements of aa.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1060≤ai≤106), where aiai is the ii-th element of aa.

Output

Print one integer — the minimum possible sum of non-deleted elements of the array after end of the game.

Examples

input

Copy

5
1 5 7 8 2

output

Copy

0

input

Copy

6
5 1 2 4 6 3

output

Copy

0

input

Copy

2
1000000 1000000

output

Copy

1000000

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + ;
int N;
int num[maxn];
vector<int> odd, even;

int main() {
    scanf("%d", &N);
    for(int i = ; i < N; i ++) {
        scanf("%d", &num[i]);
        if(num[i] % ) odd.push_back(num[i]);
        else even.push_back(num[i]);
    }
    int ans = ;
    sort(odd.rbegin(), odd.rend());
    sort(even.rbegin(), even.rend());
    int osz = odd.size(), esz = even.size();
    if(osz > esz + ) {
        for(int i = esz + ; i < osz; i ++)
            ans += odd[i];
    }
    else if(osz == esz + ) ans = ;
    else {
        for(int i = osz + ; i < esz; i ++)
            ans += even[i];
    }
    printf("%d\n", ans);
    return ;
}

C.Two Shuffled Sequences

output

standard output

Two integer sequences existed initially — one of them was strictly increasing, and the other one — strictly decreasing.

Strictly increasing sequence is a sequence of integers [x1<x2<⋯<xk][x1<x2<⋯<xk]. And strictly decreasing sequence is a sequence of integers [y1>y2>⋯>yl][y1>y2>⋯>yl]. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

They were merged into one sequence aa. After that sequence aa got shuffled. For example, some of the possible resulting sequences aa for an increasing sequence [1,3,4][1,3,4] and a decreasing sequence [10,4,2][10,4,2] are sequences [1,2,3,4,4,10][1,2,3,4,4,10] or [4,2,1,10,4,3][4,2,1,10,4,3].

This shuffled sequence aa is given in the input.

Your task is to find any two suitable initial sequences. One of them should be strictly increasing and the other one — strictly decreasing. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

If there is a contradiction in the input and it is impossible to split the given sequence aa to increasing and decreasing sequences, print "NO".

Input

The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of elements in aa.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤2⋅1050≤ai≤2⋅105), where aiai is the ii-th element of aa.

Output

If there is a contradiction in the input and it is impossible to split the given sequence aa to increasing and decreasing sequences, print "NO" in the first line.

Otherwise print "YES" in the first line and any two suitable sequences. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

In the second line print nini — the number of elements in the strictly increasing sequence. nini can be zero, in this case the increasing sequence is empty.

In the third line print nini integers inc1,inc2,…,incniinc1,inc2,…,incni in the increasing order of its values (inc1<inc2<⋯<incniinc1<inc2<⋯<incni) — the strictly increasing sequence itself. You can keep this line empty if ni=0ni=0 (or just print the empty line).

In the fourth line print ndnd — the number of elements in the strictly decreasing sequence. ndnd can be zero, in this case the decreasing sequence is empty.

In the fifth line print ndnd integers dec1,dec2,…,decnddec1,dec2,…,decnd in the decreasing order of its values (dec1>dec2>⋯>decnddec1>dec2>⋯>decnd) — the strictly decreasing sequence itself. You can keep this line empty if nd=0nd=0 (or just print the empty line).

ni+ndni+nd should be equal to nn and the union of printed sequences should be a permutation of the given sequence (in case of "YES" answer).

Examples

input

Copy

7
7 2 7 3 3 1 4

output

Copy

YES
2
3 7
5
7 4 3 2 1

input

Copy

5
4 3 1 5 3

output

Copy

YES
1
3
4
5 4 3 1

input

Copy

5
1 1 2 1 2

output

Copy

NO

input

Copy

5
0 1 2 3 4

output

Copy

YES
0

5
4 3 2 1 0 

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 2e5 + ;
int N;
int a[maxn];
int vis[maxn];

int main() {
    scanf("%d", &N);
    bool flag = true;
    for(int i = ; i < N; i ++) {
        scanf("%d", &a[i]);
        vis[a[i]] ++;
        if(vis[a[i]] > ) flag = false;
    }
    if(!flag) {
        printf("NO\n");
        return ;
    }
    sort(a, a + N);
    vector<int> up, down;
    memset(vis, , sizeof(vis));
    for(int i = ; i < N; i ++) {
        if(!vis[a[i]]) {
            up.push_back(a[i]);
            vis[a[i]] = ;
        } else down.push_back(a[i]);
    }
    printf("YES\n");
    printf("%d\n", down.size());
    if(down.size() == ) printf("\n");
    else {
        sort(down.begin(), down.end());
        for(int i = ; i < down.size(); i ++)
            printf("%d%s", down[i], i != down.size() -  ? " " : "\n");
    }
    printf("%d\n", up.size());
    if(up.size() == ) printf("\n");
    else {
        sort(up.rbegin(), up.rend());
        for(int i = ; i < up.size(); i ++)
            printf("%d%s", up[i], i != up.size() -  ? " " : "\n");
    }
    return ;
}

D. Equalize Them All

output

standard output

You are given an array aa consisting of nn integers. You can perform the following operations arbitrary number of times (possibly, zero):

1. Choose a pair of indices (i,j)(i,j) such that |i−j|=1|i−j|=1 (indices ii and jj are adjacent) and set ai:=ai+|ai−aj|ai:=ai+|ai−aj|;
2. Choose a pair of indices (i,j)(i,j) such that |i−j|=1|i−j|=1 (indices ii and jj are adjacent) and set ai:=ai−|ai−aj|ai:=ai−|ai−aj|.

The value |x||x| means the absolute value of xx. For example, |4|=4|4|=4, |−3|=3|−3|=3.

Your task is to find the minimum number of operations required to obtain the array of equal elements and print the order of operations to do it.

It is guaranteed that you always can obtain the array of equal elements using such operations.

Note that after each operation each element of the current array should not exceed 10181018 by absolute value.

Input

The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of elements in aa.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤2⋅1050≤ai≤2⋅105), where aiai is the ii-th element of aa.

Output

In the first line print one integer kk — the minimum number of operations required to obtain the array of equal elements.

In the next kk lines print operations itself. The pp-th operation should be printed as a triple of integers (tp,ip,jp)(tp,ip,jp), where tptp is either 11 or 22 (11means that you perform the operation of the first type, and 22 means that you perform the operation of the second type), and ipip and jpjp are indices of adjacent elements of the array such that 1≤ip,jp≤n1≤ip,jp≤n, |ip−jp|=1|ip−jp|=1. See the examples for better understanding.

Note that after each operation each element of the current array should not exceed 10181018 by absolute value.

If there are many possible answers, you can print any.

Examples

input

Copy

5
2 4 6 6 6

output

Copy

2
1 2 3
1 1 2

input

Copy

3
2 8 10

output

Copy

2
2 2 1
2 3 2

input

Copy

4
1 1 1 1

output

Copy

0

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 2e5 + ;
int N;
int a[maxn], vis[maxn];
int st, en, maxx, cnt = ;

int main() {
    scanf("%d", &N);
    for(int i = ; i <= N; i ++) {
        scanf("%d", &a[i]);
        vis[a[i]] ++;
        if(vis[a[i]] > cnt) {
            cnt = vis[a[i]];
            maxx = a[i];
        }
    }

    for(int i = ; i < N; i ++) {
        if(a[i] == maxx) {
            st = i;
            break;
        }
    }

    printf("%d\n", N - cnt);
    if(N - cnt == ) return ;
    for(int i = st; i >= ; i --) {
        if(a[i] == maxx) continue;
        else if(a[i] > maxx)
            printf("2 %d %d\n", i, i + );
        else printf("1 %d %d\n", i, i + );
    }
    for(int i = st; i <= N; i ++) {
        if(a[i] == maxx) continue;
        else if(a[i] < maxx)
            printf("1 %d %d\n", i, i - );
        else printf("2 %d %d\n", i, i - );
    }

    return ;
}

E. Median String

You are given two strings ss and tt, both consisting of exactly kk lowercase Latin letters, ss is lexicographically less than tt.

Let's consider list of all strings consisting of exactly kk lowercase Latin letters, lexicographically not less than ss and not greater than tt(including ss and tt) in lexicographical order. For example, for k=2k=2, s=s="az" and t=t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].

Your task is to print the median (the middle element) of this list. For the example above this will be "bc".

It is guaranteed that there is an odd number of strings lexicographically not less than ss and not greater than tt.

Input

The first line of the input contains one integer kk (1≤k≤2⋅1051≤k≤2⋅105) — the length of strings.

The second line of the input contains one string ss consisting of exactly kk lowercase Latin letters.

The third line of the input contains one string tt consisting of exactly kk lowercase Latin letters.

It is guaranteed that ss is lexicographically less than tt.

It is guaranteed that there is an odd number of strings lexicographically not less than ss and not greater than tt.

Output

Print one string consisting exactly of kk lowercase Latin letters — the median (the middle element) of list of strings of length kklexicographically not less than ss and not greater than tt.

Examples

input

Copy

2
az
bf

output

Copy

bc

input

Copy

5
afogk
asdji

output

Copy

alvuw

input

Copy

6
nijfvj
tvqhwp

output

Copy

qoztvz

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 2e5 + ;
int N;
string s, t;
int ans[maxn];

/*string add(string a, string b) {
    int len = a.length();
    int upup = 0;
    string sum = "";
    for(int i = len - 1; i >= 0; i --) {
        int na = a[i] - 'a';
        int nb = b[i] - 'a';
        int num = (na + nb) + upup;
        if(num > 25) {
            upup = 1;
            num -= 26;
        } else upup = 0;
        sum += num + 'a';
    }
    if(upup) sum = "a" + sum;
    return sum;
}*/

int main() {
    scanf("%d", &N);
    cin >> s >> t;
    for(int i = N - ; i >= ; i --) {
        int numa = s[i] - 'a';
        int numb = t[i] - 'a';
        ans[i] = numa + numb;
        if(ans[i] % ) {
            ans[i + ] += ;
            ans[i] = (ans[i] - ) / ;
            ans[i] += ans[i + ] / ;
            ans[i + ] %= ;
        }
        else ans[i] = ans[i] / ;
    }

    for(int i = ; i < N; i ++)
        printf("%c", 'a' + ans[i]);
    printf("\n");

    return ;
}

模拟 26 进制加法

F. Graph Without Long Directed Paths

You are given a connected undirected graph consisting of nn vertices and mm edges. There are no self-loops or multiple edges in the given graph.

You have to direct its edges in such a way that the obtained directed graph does not contain any paths of length two or greater (where the length of path is denoted as the number of traversed edges).

Input

The first line contains two integer numbers nn and mm (2≤n≤2⋅1052≤n≤2⋅105, n−1≤m≤2⋅105n−1≤m≤2⋅105) — the number of vertices and edges, respectively.

The following mm lines contain edges: edge ii is given as a pair of vertices uiui, vivi (1≤ui,vi≤n1≤ui,vi≤n, ui≠viui≠vi). There are no multiple edges in the given graph, i. e. for each pair (ui,viui,vi) there are no other pairs (ui,viui,vi) and (vi,uivi,ui) in the list of edges. It is also guaranteed that the given graph is connected (there is a path between any pair of vertex in the given graph).

Output

If it is impossible to direct edges of the given graph in such a way that the obtained directed graph does not contain paths of length at least two, print "NO" in the first line.

Otherwise print "YES" in the first line, and then print any suitable orientation of edges: a binary string (the string consisting only of '0' and '1') of length mm. The ii-th element of this string should be '0' if the ii-th edge of the graph should be directed from uiui to vivi, and '1' otherwise. Edges are numbered in the order they are given in the input.

Example

input

Copy

6 5
1 5
2 1
1 4
3 1
6 1

output

Copy

YES
10100

Note

The picture corresponding to the first example:

And one of possible answers:

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 2e5 + ;
int N, M;
int col[maxn], st[maxn], en[maxn];
vector<int> v[maxn];
bool flag = true;

void dfs(int st, int fa, int color) {
    col[st] = color;
    if(v[st].size() == ) return;
    for(int i = ; i < v[st].size(); i ++) {
        if(v[st][i] == fa) continue;
        if(col[v[st][i]] == -)
            dfs(v[st][i], st,  - color);
        else if(col[v[st][i]] == col[st])
            flag = false;
    }
}

int main() {
    scanf("%d%d", &N, &M);
    memset(col, -, sizeof(col));
    for(int i = ; i < M; i ++) {
        scanf("%d%d", &st[i], &en[i]);
        v[st[i]].push_back(en[i]);
        v[en[i]].push_back(st[i]);
    }
    flag = true;
    dfs(, -, );
    if(!flag) printf("NO\n");
    else {
        printf("YES\n");
        for(int i = ; i < M; i ++) {
            if(col[st[i]])
                printf("");
            else printf("");
        }
    }

    return ;
}