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: Accepted)

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

int c[26];
char s[101];

int main()
{
	int h,w;
	scanf("%d%d",&h,&w);
	for (int i=1;i<=h;i++)
	{
		scanf("%s",s+1);
		for (int j=1;j<=w;j++)
			c[s[j]-'a']++;
	}
	int c2=0,c1=0;
	for (int i=0;i<26;i++)
	{
		switch (c[i]%4)
		{
		case 1:
			c1++;
			break;
		case 2:
			c2++;
			break;
		case 3:
			c1++,c2++;
			break;
		default:
			break;
		}
	}
	if (h&1&&w&1)
	{
		if (c1==1&&c2<=(h/2)+(w/2)) puts("Yes");
		else puts("No");
	}
	else if (h&1)
	{
		if (!c1&&c2<=w/2) puts("Yes");
		else puts("No");
	}
	else if (w&1)
	{
		if (!c1&&c2<=h/2) puts("Yes");
		else puts("No");
	}
	else
	{
		if (!c1&&!c2) puts("Yes");
		else puts("No");
	}
	return 0;
}