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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
int main() {
  int H, W;
  cin >> H >> W;
  vector<int> A(50);
  REP(i, H) REP(j, W) {
    char c;
    cin >> c;
    A[c - 'a']++;
    A[c - 'a'] %= 4;
  }
  int g = 0, k = 0;
  if (W % 2 == 1) k += H / 2;
  if (H % 2 == 1) k += W / 2;
  if (W % 2 == 1 && H % 2 == 1) {
    g = 1;
  }
  REP(i, 50) {
    int a = A[i];
    if (a % 2 == 1) {
      a--;
      g--;
    }
    if (a == 2) {
      a -= 2;
      k--;
    }
  }
  if (g == 0 && k == 0) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
}