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 <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <map>

using namespace std;

int main(){
    int h, w;
    cin >> h >> w;
    vector<string> a(h);
    for(int i = 0; i < h; ++i){
        cin >> a[i];
    }
    map<char, int> cnt;
    for(int i = 0; i < h; ++i){
        for(int j = 0; j < w; ++j){
            cnt[a[i][j]]++;
        }
    }
    
    vector<int> g(3, 0);
    g[2] = (h / 2) * (w / 2);
    if(h % 2 != 0 && w % 2 != 0){
        g[0] = 1;
        g[1] = h / 2 + w / 2;
    }
    else{
        g[1] = (h * w - g[2] * 4) / 2;
    }

    vector<int> mod(3);
    mod[0] = 1;
    mod[1] = 2;
    mod[2] = 4;
    
    for(int i = 0; i < 3; ++i){
        for(auto it = cnt.begin(); it != cnt.end(); ++it){
            if(g[i] == 0)   break;
            
            int temp = it->second;
            if(temp % 4 == mod[i] || temp % 4 == 4 - mod[i]){
                it->second -= mod[i];
                --g[i];
            }
        }
    }
    
    for(auto it = cnt.begin(); it != cnt.end(); ++it){
        if(it->second != 0){
            cout << "No\n";
            return 0;
        }
    }
    cout << "Yes\n";
}