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;
typedef long long LL;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()

int main() {
  int h, w;
  cin >> h >> w;
  vector<int> num(26, 0);
  rep(i, h) rep(j, w) {
    char t; cin >> t;
    num[t - 'a'] ++;
  }

  int g1, g2, g3;
  if (h % 2 == 0 && w % 2 == 0) {
    g1 = 0, g2 = 0, g3 = h * w / 4;
  } else if (h % 2 == 1 && w % 2 == 1) {
    g1 = 1, g2 = (h / 2) + (w / 2), g3 = (h / 2) * (w / 2);
  } else {
    g1 = 0, g2 = w / 2, g3 = w * (h - 1) / 4;
  }
  

  bool check = true;
  int j = 0;
  rep(i, 26) {
    if (j == g1) break;
    if (num[i] % 4 == 1 || num[i] % 4 == 3) {
      j ++;
      num[i] --;
    }
  }
  if (j != g1) check = false;
  j = 0;
  rep(i, 26) {
    if (j == g2) break;
    if (num[i] % 4 == 2) {
      j ++;
      num[i] -= 2;
    }
  }
  if (j != g2) check = false;
  j = 0;
  rep(i, 26) {
    if (j == g3) break;
    if (num[i] > 0 && num[i] % 4 == 0) {
      j ++;
      num[i] -= 4;
    }
  }
  if (j != g3) check = false;

  rep(i, 26) {
    if (num[i] > 0) check = false;
  }

  if (check) cout << "Yes" << endl;
  else cout << "No" << endl;
}