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(void)
{
  int i, j;
  int h, w;
  char a[101];

  int ans;

  int count[26];
  int div4, div2, div1;

  scanf("%d", &h);
  scanf("%d", &w);

  for (i = 0; i < 26; i++)count[i] = 0;
  for (i = 0; i < h; i++) {
    scanf("%s", &a);
    for (j = 0; j < w; j++) {
      count[a[j] - 'a']++;
    }
  }

  div4 = div2 = div1 = 0;
  for (i = 0; i < 26; i++) {
    if (count[i] > 0) {
      if (count[i] % 4 == 0)div4++;
      else if (count[i] % 2 == 0)div2++;
      else div1++;
    }
  }

  if ((h % 2 == 0) && (w % 2 == 0)) {
    ans = ((div2 == 0) && (div1 == 0));
  } else if ((h % 2 == 1) && (w % 2 == 0)) {
    ans = ((div2 <= w) && (div1 == 0));
  } else if ((h % 2 == 0) && (w % 2 == 1)) {
    ans = ((div2 <= h) && (div1 == 0));
  } else {
    ans = ((div2 <= h + w - 2) && (div1 == 1));
  }

  printf(ans ? "Yes\n" : "No\n");

  return(0);
}