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>
int main()
{
    int h, w;
    scanf("%d %d", &h, &w);
    int i, j;
    char a[102][102];
    for (i = 0; i < h; i++)
        scanf("%s", a[i]);
    int count[30];
    for (i = 0; i < 30; i++)
        count[i] = 0;
    for (i = 0; i < h; i++)
        for (j = 0; j < w; j++)
            count[(int)(a[i][j] - 'a')]++;
    int x, y, z;
    x = y = z = 0;
    for (i = 0; i < 30; i++)
    {
        x += count[i] - (count[i] % 4);
        count[i] %= 4;
        y += count[i] - (count[i] % 2);
        z += count[i] % 2;
    }
    if (h > 1 && w > 1)
    {
        if (h % 2 == 0 && w % 2 == 0)
        {
            if (y + z > 0)
                printf("No\n");
            else
                printf("Yes\n");
        }
        else if (h % 2 == 0 && w % 2 == 1)
        {
            if (z > 0)
                printf("No\n");
            else if (y > h)
                printf("No\n");
            else
                printf("Yes\n");
        }
        else if (h % 2 == 1 && w % 2 == 0)
        {
            if (z > 0)
                printf("No\n");
            else if (y > w)
                printf("No\n");
            else
                printf("Yes\n");
        }
        else
        {
            if (z != 1)
                printf("No\n");
            else
            {
                int k = h / 2 + w / 2;
                if (y > k)
                    printf("No\n");
                else
                    printf("Yes\n");
            }
        }
    }
    else if (w % 2 == 0 || h % 2 == 0)
    {
        if (z > 0)
            printf("No\n");
        else
            printf("Yes\n");
    }
    else
    {
        if (z != 1)
            printf("No\n");
        else
            printf("Yes\n");
    }
    return 0;
}