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 <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <map>
using namespace std;
int n, m, sum;
char s[105][105];
map <string, int> p;
int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i ++) {
		cin >> s[i];
		string t = s[i];
		int l = t.size();
		sort(s[i], s[i] + l);
		//s[i] = t;
	}
	for (int i = 1; i <= n; i ++)
		p[s[i]] ++;
	for (int i = 1; i <= n; i ++) {
		if (p[s[i]] % 2 == 1) sum ++;
		if (sum > 1) {cout << "No"; return 0;}
	}
	for (int i = 1; i <= n; i ++) {
		map <char, int> p;
		for (int j = 1; j <= m; j ++)
			p[s[i][j]] ++;
		for (int j = 1; j <= m; j ++) {
			if (p[s[i][j]] % 2 == 0) sum ++;
			if (sum > 1) {cout << "No"; return 0;}
		}
	}
	cout << "Yes";
	return 0;
}