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, flag = 0;
	cin >> h >> w;
	vector<int> arr(26, 0);
	char c;
	for(int i = 0; i < h * w; i++) {
		cin >> c;
		arr[c - 'a']++;
	}
	
	if((h % 2 == 0)&&(w % 2 == 0)) {
		for(int i = 0; i < 26; i++) {
			if(arr[i] % 4 != 0) {
				flag = 1;
				break;
			}
		}
	} else if((h % 2 == 0)||(w % 2 == 0)){
		int odd, even, count;
		if(h % 2) {
			odd = h;
			even = w;
		} else {
			odd = w;
			even = h;
		}
		count = even / 2;
		for(int i = 0; i < 26; i++) {
			if(arr[i] % 4 != 0) {
				if((arr[i] % 2 == 0)&&(count)) {
					count--;
				} else {
					flag = 1;
					break;
				}
			}
		}
	} else if(h == w) {
		int one = 1, count = h - 1;
		for(int i = 0; i < 26; i++) {
			if(arr[i] % 4 != 0) {
				if(arr[i] % 2 == 1) {
					if(one == 0) {
						flag = 1;
						break;
					} else {
						one = 0;
						if((arr[i] - 1) % 4 != 0) {
							if(count != 0) {
								count--;
							} else {
								flag = 1;
								break;
							}
						}
					}
				} else {
					if(count != 0) {
						count--;
					} else {
						flag = 1;
						break;
					}
				}
			}
		}
	} else {
		int l, s, one = 1, count;
		l = max(h, w);
		s = min(h, w);
		count = l / 2;
		for(int i = 0; i < 26; i++) {
			if(arr[i] % 4 != 0) {
				if(arr[i] % 2 == 1) {
					if(one == 1) {
						if(((arr[i] - 1) % 4 != 0)&&(count != 0)) {
							count--;
						} else if(((arr[i] - 1) % 4 != 0)&&(count == 0)) {
							flag = 1;
							break;
						}
						one = 0;
					} else {
						flag = 1;
						break;
					}
				} else if(arr[i] % 2 == 0) {
					if(count == 0) {
						flag = 1;
						break;
					} else {
						count--;
					}
				}
			}
		}
	}
	if(flag) {
		cout << "No" << endl;
	} else {
		cout << "Yes" << endl;
	}

	return 0;
}