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>
#define FOR(i,m,n) for(int i=m;i<(n);i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;

int main(void){
  int H, W;
  cin >> H >> W;
  map<char, int> mojis;
  REP(i, H) REP(j, W) {
    char ch;
    cin >> ch;
    mojis[ch]++;
  }
  priority_queue<int> kazus;
  for(const auto& moji : mojis) {
    kazus.push(moji.second);
  }
  bool isW = W < H;
  while (!kazus.empty()) {
    int top = kazus.top();
    kazus.pop();
    //cerr << H << "," << W << ":" << top << endl;
    if (W > 1 || H > 1) {
      if (isW) {
        //cerr << "W < H" << endl;
        if (top < 2 * W) {
          cout << "No" << endl;
          goto end;
        }
        top -= 2 * W;
        H -= 2;
      } else {
        //cerr << "W >= H" << endl;
        if (top < 2 * H) {
          cout << "No" << endl;
          goto end;
        }
        top -= 2 * H;
        W -= 2;
      }
    } else {
      //cerr << "Last" << endl;
      if (top != 1) {
        cout << "No" << endl;
        goto end;
      }
      top = 0;
    }
    isW = W < H;
    if (top > 0)
      kazus.push(top);
  }
  cout << "Yes" << endl;
  end:
  return 0;
}