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

int main() {
	int h, w, a[26] = {};
	cin >> h >> w;
	for (int i = 0; i < h; i++) {
		string s;
		cin >> s;
		for (int j = 0; j < w; j++) {
			a[s[j] - 'a']++;
		}
	}
	int k1 = 0, k2 = 0;
	for (int i = 0; i < 26; i++) {
		if (a[i] % 2) {
			k1++;
		} else if (a[i] % 4) {
			k2++;
		}
	}
	int m = min(h, w);
	int n = max(h, w);
	if (m % 2 && n % 2) {
		cout << (k1 == 1 && k2 >= (n - m) / 2 && k2 <= (m + n - 2) / 2 ? "Yes" : "No") << endl;
	} else if (m % 2) {
		cout << (k1 == 0 && k2 <= n / 2 ? "Yes" : "No") << endl;
	} else if (n % 2) {
		cout << (k1 == 0 && k2 <= m / 2 ? "Yes" : "No") << endl;
	} else {
		cout << (k1 == 0 && k2 == 0 ? "Yes" : "No") << endl;
	}
	return 0;
}