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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <numeric>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

int main() {
    int h, w;
    cin >> h >> w;

    int p[26] = {};
    for (int i = 0; i < h * w; i++) {
        char a;
        cin >> a;
        a -= 'a';

        p[a]++;
    }

    int x = 0;
    for (int i = 0; i < 26; i++) {
        x += p[i] % 4;
    }
    int y = 0;
    for (int i = 0; i < 26; i++) {
        y += p[i] % 2;
    }

    int b = 0;
    if (h % 2 == 0 && w % 2 == 0) {
        b = x == 0;
    } else if (h % 2 == 0 && w % 2 != 0) {
        b = x <= h && y == 0;
    } else if (h % 2 != 0 && w % 2 == 0) {
        b = x <= w && y == 0;
    } else {
        b = x <= h + w - 1 && y == 1;
    }

    cout << (b ? "Yes" : "No") << endl;

    return 0;
}