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: Accepted)

#include <bits/stdc++.h>
using namespace std;

int h, w;
priority_queue<int> pq;
vector<int> cnt(26, 0);

int main() {
  cin >> h >> w;
  for (int i = 0; i < h; ++i) {
    string s;
    cin >> s;
    for (auto c : s) ++cnt[c - 'a'];
  }
  for (int i = 0; i < 26; ++i) pq.push(cnt[i]);
  vector<int> v;
  for (int i = 0; i < (h + 1) / 2; ++i)
    for (int j = 0; j < (w + 1) / 2; ++j) {
      int now = 4;
      if ((h & 1) && i == h / 2) now >>= 1;
      if ((w & 1) && j == w / 2) now >>= 1;
      v.push_back(now);
    }
  sort(v.begin(), v.end(), greater<int>());
  for (auto p : v) {
    if (p > pq.top()) {
      cout << "No" << endl;
      return 0;
    }
    p = pq.top() - p;
    pq.pop();
    pq.push(p);
  }
  cout << "Yes" << endl;
  return 0;
}