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 <map>
using namespace std;
    
int n,m;

bool column(vector<vector<char>> strs){
    map<char, int> map;
    for(int i=0; i<n; i++){
        for(int l=0; l<m; l++){
            map[strs[i][l]] += 1;
        }
        bool already=false;
        for(auto x: map){
            if(x.second%2 != 0){
                if(already == false){
                    already = true;
                }else{
                    return false;
                }
            }
        }
        map.clear();
    }
    return true;
}

bool row(vector<vector<char>> strs){
    map<char, int> map;
    for(int i=0; i<m; i++){
        for(int l=0; l<n; l++){
            map[strs[l][i]] += 1;
        }
        bool already=false;
        for(auto x: map){
            if(x.second%2 != 0){
                if(already == false){
                    already = true;
                }else{
                    return false;
                }
            }
        }
        map.clear();
    }
    return true;
}

int main() {
    cin >> n >> m;
    vector<vector<char>> map(n);
    for(int i=0; i<n; i++){
        map[i].reserve(m);
        for(int l=0; l<m; l++){
            char a; cin >> a;
            map[i].push_back(a);
        }
    }
    if(row(map) && column(map)){
        cout << "Yes" << endl;
    }else{
        cout << "No" << endl;
    }
}