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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#define CHARACTERS 26

int make_palindrome(int*, const int, const int);

int main(void){
    int char_count[CHARACTERS] = {0};
    int H, W;
    int i, j;
    char character;

    scanf("%d %d", &H, &W);
    for(i=0; i<H; i++){
        scanf(" %c", &character);
        char_count[(int)character - 'a']++;
        for(j=1; j<W; j++){
            scanf("%c", &character);
            char_count[(int)character - 'a']++;
        }
    }

    for(i=0; i<H/2; i++){
        if(!make_palindrome(char_count, W, 2)){
            printf("No");
            return 0;
        }
    }
    if(H%2 == 1){
        if(!make_palindrome(char_count, W, 1)){
            printf("No");
            return 0;
        }
    }

    printf("Yes");
    return 0;
}

//n個の回文(長さsize)が作れるかどうかチェック 1:作れる 0:作れない
int make_palindrome(int *array, const int size, const int n){
    int i, j;

    //列数が奇数の場合
    if(size%2 == 1){
        for(i=0; i<size/2; i++){
            for(j=0; j<CHARACTERS; j++){
                if(array[j] >= 2*n){
                    array[j] -= 2*n;
                    break;
                }else if(j == CHARACTERS-1){
                    return 0;
                }
            }
        }
        for(i=0; i<CHARACTERS; i++){
                if(array[i] >= 1*n){
                    array[i] -= 1*n;
                    break;
                }else if(i == CHARACTERS-1){
                    return 0;
                }
        }
        return 1;
    }

    //列数が偶数の場合
    if(size%2 == 0){
        for(i=0; i<size/2; i++){
            for(j=0; j<CHARACTERS; j++){
                if(array[j] >= 2*n){
                    array[j] -= 2*n;
                    break;
                }else if(j == CHARACTERS-1){
                    return 0;
                }
            }
        }
        return 1;
    }

    return 1;
}