Codeforces Round #381 (Div. 2) A B C 水 构造

A. Alyona and copybooks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.

What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.

Input

The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).

Output

Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.

Examples

input

1 1 3 4

output

3

input

6 2 1 1

output

1

input

4 4 4 4

output

0

input

999999999 1000000000 1000000000 1000000000

output

1000000000

Note

In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.

In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.

In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.

In the fourth example Alyona should buy one pack of one copybook.

题意：1件的费用为a  2件的费用为b  3件的费用为c  现在已经拥有n件物品  购买另外k件 使得(n+k)%4==0 输出最小的花费

题解；水 共有9种策略

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #define ll __int64
 #define mod 10000000007
 using namespace std;
 ll n,a,b,c;
 int main()
 {
     scanf("%I64d %I64d %I64d %I64d",&n,&a,&b,&c);
     if(n%==)
     {
         printf("0\n");
     }
     else
     {
         int exm=-n%;
         if(exm==)
         {
             printf("%I64d\n",min(a,min(b+c,*c)));
         }
         if(exm==)
         {
             printf("%I64d\n",min(*a,min(b,c*)));
         }
         if(exm==)
         {
             printf("%I64d\n",min(*a,min(b+a,c)));
         }
     }
     return ;
 }

B. Alyona and flowers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.

Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.

For example, consider the case when the mother has 5 flowers, and their moods are equal to 1,  - 2, 1, 3,  - 4. Suppose the mother suggested subarrays (1,  - 2), (3,  - 4), (1, 3), (1,  - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:

• the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
• the second flower adds ( - 2)·1 =  - 2, because he is in one of chosen subarrays,
• the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
• the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
• the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.

Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!

Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.

The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).

The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri(1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].

Each subarray can encounter more than once.

Output

Print single integer — the maximum possible value added to the Alyona's happiness.

Examples

input

5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4

output

7

input

4 3
1 2 3 4
1 3
2 4
1 1

output

16

input

2 2
-1 -2
1 1
1 2

output

0

Note

The first example is the situation described in the statements.

In the second example Alyona should choose all subarrays.

The third example has answer 0 because Alyona can choose none of the subarrays.

题意：给你n个数 m个区间  在这m个区间中取若干个区间  对于选择的区间求和累加 输出最大值

题解：水 对于某个区间 若区间和大于0则累加否则舍弃

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #define ll __int64
 #define mod 10000000007
 using namespace std;
 int n,m;
 int a[];
 int sum[];
 int b[];
 int l,r;
 int  main()
 {
     scanf("%d %d",&n,&m);
     sum[]=;
     for(int i=;i<=n;i++)
     {
         scanf("%d",&a[i]);
         sum[i]=sum[i-]+a[i];
     }
     for(int i=;i<=m;i++)
     {
         scanf("%d %d",&l,&r);
         b[i]=sum[r]-sum[l-];
     }
     sort(b+,b++m);
     int ans=;
     for(int i=m;i>=;i--)
         ans=ans+max(,b[i]);
     printf("%d\n",ans);
     return ;
 }

C. Alyona and mex

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Alyona's mother wants to present an array of n non-negative integers to Alyona. The array should be special.

Alyona is a capricious girl so after she gets the array, she inspects m of its subarrays. Subarray is a set of some subsequent elements of the array. The i-th subarray is described with two integers li and ri, and its elements are a[li], a[li + 1], ..., a[ri].

Alyona is going to find mex for each of the chosen subarrays. Among these m mexes the girl is going to find the smallest. She wants this minimum mex to be as large as possible.

You are to find an array a of n elements so that the minimum mex among those chosen by Alyona subarrays is as large as possible.

The mex of a set S is a minimum possible non-negative integer that is not in S.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105).

The next m lines contain information about the subarrays chosen by Alyona. The i-th of these lines contains two integers li and ri(1 ≤ li ≤ ri ≤ n), that describe the subarray a[li], a[li + 1], ..., a[ri].

Output

In the first line print single integer — the maximum possible minimum mex.

In the second line print n integers — the array a. All the elements in a should be between 0 and 109.

It is guaranteed that there is an optimal answer in which all the elements in a are between 0 and 109.

If there are multiple solutions, print any of them.

Examples

input

5 3
1 3
2 5
4 5

output

2
1 0 2 1 0

input

4 2
1 4
2 4

output

3
5 2 0 1

Note

The first example: the mex of the subarray (1, 3) is equal to 3, the mex of the subarray (2, 5) is equal to 3, the mex of the subarray (4, 5)is equal to 2 as well, thus the minumal mex among the subarrays chosen by Alyona is equal to 2.

题意：要求你构造一个长度为n的数列  给你m个区间  某个区间的mex值=当前区间内没有出现过的最小的非负数  要求构造的数列中 区间mex的最小值尽可能大

题解：找到给定区间的最小长度 len   构造一个 0 1 2...len-1 0 1 2...len-1...的数列 满足题目要求  并且mex值的最小值为len

为说明问题 对于题目的样例一

5 3
1 3
2 5
4 5
得出的结果为
0 1 0 1 0   
试验证

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #define ll __int64
 #define mod 10000000007
 using namespace std;
 int n,m;
 int ans[];
 int l[],r[];
 int main()
 {
     scanf("%d %d",&n,&m);
     int exm=;
     for(int i=;i<=m;i++)
     {
         scanf("%d %d",&l[i],&r[i]);
         exm=min(exm,r[i]-l[i]);
     }
     for(int i=;i<=n;i++)
        ans[i]=;
     for(int i=;i<=n;i++)
     {
       for(int j=;j<=exm;j++)
       {
           if(i+j<=n)
           {
               ans[i+j]=j;
           }
       }
       i=i+exm;
     }
     printf("%d\n",exm+);
     for(int i=;i<=n;i++)
         printf("%d ",ans[i]);
     printf("\n");
     return ;
 }