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>
using namespace std;

char matrix[110][110];
char C[256];

int main() {
    int H, W;
    cin >> H >> W;

    for (int i=0; i<H; i++) cin >> matrix[i];

    for (int i=0; i<H; i++) {
        for (int j=0; j<W; j++) {
            C[matrix[i][j]]++;
        }
    }

    if (W == 1 || H == 1) {
        int remain = 0;
        for (int i=0; i<256; i++) {
            remain += C[i] % 2;
        }
        if (W % 2 && H % 2) {
            cout << (remain == 1 ? "Yes" : "No") << endl;
        } else {
            cout << (remain ? "No" : "Yes") << endl;
        }

    } else {
        for (int i=0; i<256; i++) C[i] %= 4;
        if (W % 2 == 0 && H % 2 == 0) {
            int remain = 0;
            for (int i=0; i<256; i++) remain += C[i];
            cout << (remain ? "No" : "Yes") << endl;
        } else if (W % 2 && H % 2) {
            int remain = 0;
            int used = 0;
            for (int i=0; i<256; i++) {
                remain += C[i] % 2;
                used += C[i] - (C[i] % 2);
            }
            cout << (used <= H + W - 2 && remain == 1 ? "Yes" : "No") << endl;
        } else {
            int remain = 0;
            int used = 0;
            for (int i=0; i<256; i++) {
                remain += C[i] % 2;
                used += C[i] - (C[i] % 2);
            }
            if (H % 2 == 0) {
                cout << (used <= H && remain == 0 ? "Yes" : "No") << endl;
            } else {
                cout << (used <= W && remain == 0 ? "Yes" : "No") << endl;
            }
        }
    }


    return 0;
}