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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int a[26] = {};
char x[101][101] = {};
int main()
{
	int h, w;
	int cnt_4 = 0,cnt_2 = 0,cnt_1 = 0;
	int ans = 0;
	scanf("%d %d",&h, &w);
	for (int i = 0; i < h; i++)
	{
			scanf("%s",x[i]);
			for (int j = 0; j < w; j++)
			{
				a[x[i][j] - 'a']++;
			}
	}

	for (int i = 0; i < 26; i++)
	{
		int x, y;
		x = 0;
		y = 0;
		x = a[i];
		cnt_4 = cnt_4 + x / 4;
		if (x % 4 != 0)
		{
			y = x % 4;
			cnt_2 = cnt_2 + y / 2;
		}
		if (x % 4 != 0 && y % 2 != 0)
		{
			cnt_1 = cnt_1 + y % 2;
		}
	}
	if (h % 2 != 0 && w % 2 != 0)
	{
		if (cnt_1 == 1 && cnt_4 >= (w / 2) * (h / 2))
		{
			int q;
			q = cnt_4 - (w / 2) * (h / 2);
				if (cnt_2 + (q * 4) == w / 2 + h / 2)
				{
					ans = 1;
				}
		}
	}
	else if (h % 2 == 0 && w % 2 != 0)
	{
		if (cnt_1 == 0 && cnt_4 >= (w / 2) * (h / 2))
		{
			int q;
			q = cnt_4 - (w / 2) * (h / 2);
			if (cnt_2 + (q * 4) == h / 2)
			{
				ans = 1;
			}
		}
	}
	else if (h % 2 != 0 && w % 2 == 0)
	{
		if (cnt_1 == 0 && cnt_4 >= (w / 2) * (h / 2))
		{
			int q;
			q = cnt_4 - (w / 2) * (h / 2);
			if (cnt_2 + (q * 4) == w / 2)
			{
				ans = 1;
			}
		}
	}
	else if (h % 2 == 0 && w % 2 == 0)
	{
		if (cnt_1 == 0 && cnt_2 == 0 && cnt_4 == (w / 2) * (h / 2))
		{
			ans = 1;
		}
	}
	if (ans == 1)
	{
		printf("Yes");
	}
	else
	{
		printf("No");
	}
	return 0;
}