【25.00%】【codeforces 584E】Anton and Ira

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Anton loves transforming one permutation into another one by swapping elements for money, and Ira doesn’t like paying for stupid games. Help them obtain the required permutation by paying as little money as possible.

More formally, we have two permutations, p and s of numbers from 1 to n. We can swap pi and pj, by paying |i - j| coins for it. Find and print the smallest number of coins required to obtain permutation s from permutation p. Also print the sequence of swap operations at which we obtain a solution.

Input

The first line contains a single number n (1 ≤ n ≤ 2000) — the length of the permutations.

The second line contains a sequence of n numbers from 1 to n — permutation p. Each number from 1 to n occurs exactly once in this line.

The third line contains a sequence of n numbers from 1 to n — permutation s. Each number from 1 to n occurs once in this line.

Output

In the first line print the minimum number of coins that you need to spend to transform permutation p into permutation s.

In the second line print number k (0 ≤ k ≤ 2·106) — the number of operations needed to get the solution.

In the next k lines print the operations. Each line must contain two numbers i and j (1 ≤ i, j ≤ n, i ≠ j), which means that you need to swap pi and pj.

It is guaranteed that the solution exists.

Examples

input

4

4 2 1 3

3 2 4 1

output

3

2

4 3

3 1

Note

In the first sample test we swap numbers on positions 3 and 4 and permutation p becomes 4 2 3 1. We pay |3 - 4| = 1 coins for that. On second turn we swap numbers on positions 1 and 3 and get permutation 3241 equal to s. We pay |3 - 1| = 2 coins for that. In total we pay three coins.

【题目链接】:http://codeforces.com/contest/584/problem/E

【题解】



可以把第二个排列看成是1..n的排列;

即f[s2[1]] = 1,f[s2][2]]=2…f[s2[i]] = i;

然后再把第一个排列中的各个数字用这个映射关系修改一下

即s1[1] = f[s1[1]],s1[2] = f[s1[2]]….s1[i] = f[s1[i]];

然后问题就转换成把一个无序的排列改成有序的过程.

对于每个s1[i]!=i的元素来说；它最后肯定要变成s[i]==i;

则找到s1[j]==i的下标j;

交换的代价就是|i-j|;

但是直接硬生生地这样交换肯定不行的；

swap(i,j)的代价和swap(i,k),swap(k,l),swap(l,….)….swap(..,x),swap(x,j)的总代价是一样的；

则我们完全可以把在j和i之间其他的数字也往前移动,即s[k]<=p的数字,其中p是i这个数字当前所在的位置,(显然这个位置的数字放在前面更优);

这样我们在做一件事情的时候，在不消耗多余花费的时候，尽量让答案更靠近了最优解;



【完整代码】

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
using namespace std;
const int MAXN = 2100;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);
int n;
int a[MAXN],f[MAXN];
vector < pair<int,int> > ans;
void rel(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}
void rei(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}
int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);
    for (int i = 1;i <= n;i++)
        rei(a[i]);
    for (int i = 1;i <= n;i++)
    {
        int t;
        rei(t);
        f[t] = i;
    }
    for (int i = 1;i <= n;i++)
        a[i] = f[a[i]];
    LL tot = 0;
    for (int i = n;i >= 1;i--)
        if (a[i] == i)
            continue;
        else
        {
            int pos;
            for (int j = 1;j <= n;j++)
                if (a[j] == i)
                {
                    pos = j;
                    break;
                }
            int j = pos+1;
            while (pos != i)
                if (a[j] <= pos)
                {
                    ans.push_back(make_pair(j,pos));
                    swap(a[j],a[pos]);
                    tot+=j-pos;
                    pos = j;
                    j++;
                }
                else
                    j++;
        }
    printf("%I64d\n",tot);
    int len = ans.size();
    printf("%d\n",len);
    for (int i = 0;i <= len-1;i++)
        printf("%d %d\n",ans[i].first,ans[i].second);
    return 0;
}