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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <iostream>
#include <math.h>
#include <algorithm>
#include <vector>
using namespace std;

int h, w;
int x[26];
char a;
bool ans = 1;
int flag2 = 0, flag1 = 0;

int main(void) {
  scanf("%d %d",&h, &w);
  for(int i=0;i<h*w;i++){
    cin>>a;
    x[a-97]++;
  }

  for(int i=0;i<26;i++){
    if(x[i] % 4 != 0){
      if(x[i] % 2 != 0) flag1++;
      else flag2++;
      break;
    }
  }


  if( flag1 > (h*w)%2 ) ans = 0;
  if(flag2 > ( (h%2) * w + (w%2) * h )/2 - ((h*w)%2))	ans = 0;

  if(ans) printf("Yes\n");
  else printf("No\n");

  return 0;
}