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: Accepted)

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

string a[105];

int main(){
    int h, w;
    cin >> h >> w;
    for(int i=0;i<h;++i)
        cin >> a[i];

    map<char, int> mp;
    for(int i=0;i<h;++i){
        for(int j=0;j<w;++j){
            mp[a[i][j]]++;
        }
    }

    for(char c='a';c<='z';++c){
        while(mp[c] >= 4)mp[c] -= 4;
    }
    if((h%2 != 0) || (w%2 != 0)){
        int limit;
        if((h%2 != 0) && (w%2 != 0))limit = w+h-2;
        else if(h%2 != 0)limit = w;
        else if(w%2 != 0)limit = h;
        else limit = 0;
        int sum = 0;
        for(char c='a';c<='z';++c){
            while(mp[c] >= 2 && sum <= limit){
                mp[c] -= 2;
                sum += 2;
            }
        }
    }
    if((h%2 != 0) && (w%2 != 0)){
        for(char c='a';c<='z';++c){
            if(mp[c] > 0){
                mp[c]--;
                break;
            }
        }
    }

    bool flag = true;
    for(char c='a';c<='z';++c){
        if(mp[c] > 0){
            flag = false;
            break;
        }
    }

    if(flag) cout << "Yes" << endl;
    else cout << "No" << endl;

    return 0;
}