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>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main(void){
  int h, w, i, n=0, ans[26], f[3]={0,0,0};
  char c;
  for(i=0;i<26;i++) ans[i]=0;
  scanf("%d %d",&h,&w);
  for(i=0;i<h*w;i++){
    scanf("%c",&c);
    ans[c-'a']++;
  }
  if(h==1 && w==1) printf("Yes\n");
  else{
    for(i=0;i<26;i++){
      if(ans[i]%4!=0) f[ans[i]%4-1]++;
      if(f[0]+f[2]>=2) printf("No\n");
      else{
	if(h%2==1) n+=w/2;
	if(w%2==1) n+=h/2;
	if(n<f[1]) printf("No\n");
	else if(h%2==1 && w%2==1 && (f[0]+f[2]!=1 || (n==f[1] && f[2]==1))){
	  printf("No\n");
	}else printf("Yes\n");
      }
    }
  }
  return 0;
}