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;

int main() {
	// your code goes here a=97
	long atoz[26]={0};
	long H,W;
	cin >>H>>W;
	
	for(long i=0; i<H*W; i++){
		char a;
		cin >>a ;
		long b=a-97;
		atoz[b]++;
	}			//文字数カウンタ
	
	int flag=1;
	//行・列数ともに偶数のとき
	if(H%2==0 && W%2==0){
		for(int i=0; i<26; i++){
			if(atoz[i]%4!=0){
				flag=0;
				break;
			}
		}
	}
	
	else if(H%2==0 && W%2==1){
		long yonbai=0;
		for(int i=0; i<26; i++){
			if(atoz[i]%2==1){
				flag=0;
				break;
			}
			yonbai+=atoz[i]/4;
		}
		if(yonbai < (W-1)*H/4)flag=0;
	}
	else if(H%2==1 && W%2==0){
		long yonbai=0;
		for(int i=0; i<26; i++){
			if(atoz[i]%2==1){
				flag=0;
				break;
			}
			yonbai+=atoz[i]/4;
		}
		if(yonbai < (H-1)*W/4)flag=0;
	}
	else{
		long kisu=0;
		long yonbai=0;
		for(int i=0; i<26; i++){
			if(atoz[i]%2==1)kisu++;
			if(atoz[i]%4==0)yonbai++;
		}
		if(kisu>=2)flag=0;
		if(yonbai<(H-1)*(W-1)/4)flag=0;
	}
	
	
	if(flag==0)cout<<"No"<<endl;
	else cout<<"Yes"<<endl;
	return 0;
}