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;

signed main(){	
	int h, w;
	cin >> h >> w;
	int alp[26];
	for (int i = 0;i < 26;i++)
		alp[i] = 0;
	for (int i = 0;i < h;i++){
		string ai;
		cin >> ai;
		for (int j = 0;j < w;j++){
			alp[ai[j] - 'a']++;
		}
	}
	/*
		c0 : 4の倍数文字の文字数
		c1 : 4の倍数+1文字の文字数
		c2 : 4の倍数+2文字の文字数
		c3 : 4の倍数+3文字の文字数
	*/
	int c[4];
	for (int i = 0;i < 4;i++) c[i] = 0;
	for (int i = 0;i < 26;i++)
		c[alp[i] & 3]++;
	if (((h & 1) == 0) && ((w & 1) == 0)){ // hもwも偶数
		/* alpすべて4の倍数じゃないとだめ */
		bool f = true;
		for (int i = 0;i < 4;i++){
			f = f && (c[i] & 3) == 0;
		}
		if (f){
			cout << "Yes" << endl;
		}
		else{
			cout << "No" << endl;
		}
	}
	else if (h & w & 1){ // hもwも奇数
		if (c[2] <= h / 2 + w / 2 && c[1] == 1 && c[3] == 0){
			cout << "Yes" << endl;
		}
		else if (c[2] <= h / 2 + w / 2 - 1 && c[1] == 0 && c[3] == 1){
			cout << "Yes" << endl;
		}
		else{
			cout << "No" << endl;
		}
	}
	else if (h & 1){ // hのみ奇数
		if (c[1] == 0 && c[3] == 0 && c[2] <= w / 2){
			cout << "Yes" << endl;
		}
		else{
			cout << "No" << endl;
		}
	}
	else{
		if (c[1] == 0 && c[3] == 0 && c[2] <= h / 2){
			cout << "Yes" << endl;
		}
		else{
			cout << "No" << endl;
		}
	}
	return 0;
}