Problem:
Score : 400 points

Problem Statement

We have an H-by-W matrix.
Let a_{ij} be the element at the i-th row from the top and j-th column from the left.
In this matrix, each a_{ij} is a lowercase English letter.

Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A.
Here, he wants to satisfy the following condition:

Every row and column in A' can be read as a palindrome.

Determine whether he can create a matrix satisfying the condition.

Note

A palindrome is a string that reads the same forward and backward.
For example, a, aa, abba and abcba are all palindromes, while ab, abab and abcda are not.

Constraints

1 ≤ H, W ≤ 100

a_{ij} is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}

Output

If Snuke can create a matrix satisfying the condition, print Yes; otherwise, print No.

Sample Input 1

3 4
aabb
aabb
aacc

Sample Output 1

Yes

For example, the following matrix satisfies the condition.

abba
acca
abba

Sample Input 2

2 2
aa
bb

Sample Output 2

No

It is not possible to create a matrix satisfying the condition, no matter how we rearrange the elements in A.

Sample Input 3

5 1
t
w
e
e
t

Sample Output 3

Yes

For example, the following matrix satisfies the condition.

t
e
w
e
t

Sample Input 4

2 5
abxba
abyba

Sample Output 4

No

Sample Input 5

1 1
z

Sample Output 5

Yes

Solution in C (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <limits.h>

int main(void){
    int h,w,count[26]={};
    scanf("%d%d",&h,&w);
    char a[h][w+1],lowcmp[27]="abcdefghijklmnopqrstuvwxyz";
    for (int i=0; i<h; i++){
        scanf("%s",a[i]);
    }
    for (int i=0; i<h; i++){
        for (int j=0; j<w; j++){
            for (int k=0; k<26; k++){
                if (a[i][j]==lowcmp[k]){
                    count[k]++;
                }
            }
        }
    }
    int sum=0;
    if (h%2==0 && w%2==0){
        for (int i=0; i<26; i++){
            if (count[i]%4!=0){
                printf("No\n");
                return 0;
            }
        }
        
    } else if (w%2==0){
        for (int i=0; i<26; i++){
            if (count[i]%2!=0){
                printf("No\n");
                return 0;
            } else if (count[i]%4!=0){
                sum+=count[i]%4;
                if (sum>w){
                    printf("No\n");
                    return 0;
                }
            }
        }
    } else if (h%2==0){
        for (int i=0; i<26; i++){
            if (count[i]%2!=0){
                printf("No\n");
                return 0;
            } else if (count[i]%4!=0){
                sum+=count[i]%4;
                if (sum>h){
                    printf("No\n");
                    return 0;
                }
            }
        }
    } else {
        bool odd=false;
        for (int i=0; i<h; i++){
            if (count[i]%2!=0){
                if (odd==true){
                    printf("No\n");
                    return 0;
                }
                odd=true;
            }
            if (count[i]%4!=0){
                sum+=count[i]%4;
                if (sum>h+w-1){
                    printf("No\n");
                    return 0;
                }
            }
        }
    }
    printf("Yes\n");
	return 0 ;
}