codeforces #577(Div.2)

codeforces #577(Div.2)

A  Important Exam

A class of students wrote a multiple-choice test.

There are nn students in the class. The test had mm questions, each of them had 55 possible answers (A, B, C, D or E). There is exactly one correct answer for each question. The correct answer for question ii worth aiai points. Incorrect answers are graded with zero points.

The students remember what answers they gave on the exam, but they don't know what are the correct answers. They are very optimistic, so they want to know what is the maximum possible total score of all students in the class.

Input

The first line contains integers nn and mm (1≤n,m≤10001≤n,m≤1000) — the number of students in the class and the number of questions in the test.

Each of the next nn lines contains string sisi (|si|=m|si|=m), describing an answer of the ii-th student. The jj-th character represents the student answer (A, B, C, D or E) on the jj-th question.

The last line contains mm integers a1,a2,…,ama1,a2,…,am (1≤ai≤10001≤ai≤1000) — the number of points for the correct answer for every question.

Output

Print a single integer — the maximum possible total score of the class.

Examples

input

2 4
ABCD
ABCE
1 2 3 4

output

16

input

3 3
ABC
BCD
CDE
5 4 12

output

21

Note

In the first example, one of the most optimal test answers is "ABCD", this way the total number of points will be 1616.

In the second example, one of the most optimal test answers is "CCC", this way each question will be answered by exactly one student and the total number of points is 5+4+12=215+4+12=21.

题意：学生考完试后想算成绩，但他们不知道正确答案，只是每个人都记住了自己填写的答案，最后一行输入每个题目如果正确的话其得分为多少。让你算一下答案由你来决定的话，它们总分最多是多少，输出这个总分即可。每道题目最多有五个选项。

思路：算出每个题中选项最多的答案数量，再乘以改题正确的分数。代码如下：

#include <bits/stdc++.h>
#define LL long long
using namespace std;
int const maxn=1001;
struct node{
	string a;
}q[maxn];
int main()
{
	int n,k,x;
	LL sum=0;
	int s[5];
	scanf("%d %d",&n,&k);
	for(int i=0;i<n;i++){
		cin>>q[i].a;
	}
	for(int i=0;i<k;i++){
		scanf("%d",&x);
		memset(s,0,sizeof(s));
		for(int j=0;j<n;j++){
			if(q[j].a[i]=='A')s[0]++;
			if(q[j].a[i]=='B')s[1]++;
			if(q[j].a[i]=='C')s[2]++;
			if(q[j].a[i]=='D')s[3]++;
			if(q[j].a[i]=='E')s[4]++;
		}
		sum+=x*max(s[0],max(s[1],max(s[2],max(s[3],s[4]))));
	}
	cout<<sum<<endl;
	return 0;
}

B.Zero Array

You are given an array a1,a2,…,ana1,a2,…,an.

In one operation you can choose two elements aiai and ajaj (i≠ji≠j) and decrease each of them by one.

You need to check whether it is possible to make all the elements equal to zero or not.

Input

The first line contains a single integer nn (2≤n≤1052≤n≤105) — the size of the array.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the elements of the array.

Output

Print "YES" if it is possible to make all elements zero, otherwise print "NO".

Examples

input

4
1 1 2 2

output

YES

input

6
1 2 3 4 5 6

output

NO

Note

In the first example, you can make all elements equal to zero in 33 operations:

• Decrease a1a1 and a2a2,
• Decrease a3a3 and a4a4,
• Decrease a3a3 and a4a4

In the second example, one can show that it is impossible to make all elements equal to zero.

题意： 输入一组数，每次选其中两个数，使这两个数都减一。输入的数能否全都减为0。能的话输出“YES”，不能的话输出“NO”。

思路：计算所有数的和，和为奇数肯定不行，和为偶数有可能。但当某个数比其他所有数的和都大时也不行。代码如下：

#include <bits/stdc++.h>
#define LL long long
using namespace std;
int const maxn=100001;
bool cmp(int a,int b){return a>b;}
int main(){
	int n;
	int a[maxn];
	LL sum=0;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
		sum+=a[i];
	}
	sort(a,a+n,cmp);
	if(sum%2==0&&sum-a[0]>=a[0])printf("YES\n");
	else printf("NO\n");
	return 0;
}