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<bits/stdc++.h>
using namespace std;
#ifdef LOCAL_DEBUG
  #include "LOCAL_DEBUG.hpp"
#endif
#define int long long

signed main(){

  int h, w; cin >> h >> w;
  vector<string> a(h);
  for(int i = 0; i < h; i++){
    cin >> a[i];
  }

  int one = h%2 == 1 && w%2 == 1;
  int two = (h%2 == 1 ? h/2 : 0) + (w%2 == 1 ? w/2 : 0);
  int four = h/2 * (w/2);
  vector<int> cnt(26, 0);
  for(int i = 0; i < h; i++){
    for(char c : a[i]){
      cnt[c - 'a']++;
    }
  }

  for(int i = 0; i < 26; i++){
    if(four > 0 && cnt[i] >= 4){
      four--;
      cnt[i] -= 4;
    }
  }
  for(int i = 0; i < 26; i++){
    if(two > 0 && cnt[i] >= 2){
      two--;
      cnt[i] -= 2;
    }
  }
  for(int i = 0; i < 26; i++){
    if(one > 0 && cnt[i] >= 1){
      one--;
      cnt[i] -= 1;
    }
  }
  cout << (one == 0 && two == 0 && four == 0 ? "Yes" : "No") << endl;

  return 0;
}