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<iostream>
#include<vector>
#include<algorithm>
#include<string>
using namespace std;


int main(){
    int n, m;
    cin >> n >> m;
    vector<int> a(26, 0);
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            char x;
            cin >> x;
            a[x - 'a']++;
        }
    }
    int tot = n*m;
    int a4 = (n/2)*(m/2); //cout << a4;
    tot -= a4*4;
    for(int i = 0; i < 26; i++){
        while(a[i]  >= 4 && a4 > 0){
            a[i] -= 4;
            a4--;
        }
    }
    
    if(a4) {
        cout << "No\n"; return 0;
    }

    if(tot == 0) {
        cout << "Yes\n"; return 0;
    }
    int a2 = (m/2)*(n%2) + (n/2)*(m%2);
    tot -= a2*2;
    for(int i = 0; i < 26; i++){
        while(a[i]  >= 2 && a2 > 0){
            a[i] -= 2;
            a2--;
        }
    }
    if(a2) {
        cout << "No\n"; return 0;
    }

    if(tot == 0) {
        cout << "Yes\n"; return 0;
    }
    for(int i = 0; i < 26; i++){
        if(a[i]) {
            cout << "Yes\n"; return 0;
        }
    }
    cout << "No";

}