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: Accepted)

#include <stdio.h>

int main(void) {
	int i, j, k;
	int h, w;
	char a[100][101];
	int cnt[256] = {0};

	scanf("%d %d", &h, &w);
	for (i = 0; i < h; i++)
		scanf("%s", a[i]);

	for (i = 0; i < h; i++)
		for (j = 0; j < w; j++)
			cnt[a[i][j]]++;

	for (i = 0; i < h / 2; i++)
		for (j = 0; j < w / 2; j++) {
			for (k = 'a'; k <= 'z'; k++)
				if (cnt[k] >= 4) {
					cnt[k] -= 4;
					break;
				}
			if (k > 'z') {
				puts("No");
				return 0;
			}
		}

	for (i = h / 2; i < (h + 1) / 2; i++)
		for (j = 0; j < w / 2; j++) {
			for (k = 'a'; k <= 'z'; k++)
				if (cnt[k] >= 2) {
					cnt[k] -= 2;
					break;
				}
			if (k > 'z') {
				puts("No");
				return 0;
			}
		}

	for (i = 0; i < h / 2; i++)
		for (j = w / 2; j < (w + 1) / 2; j++) {
			for (k = 'a'; k <= 'z'; k++)
				if (cnt[k] >= 2) {
					cnt[k] -= 2;
					break;
				}
			if (k > 'z') {
				puts("No");
				return 0;
			}
		}

	for (i = h / 2; i < (h + 1) / 2; i++)
		for (j = w / 2; j < (w + 1) / 2; j++) {
			for (k = 'a'; k <= 'z'; k++)
				if (cnt[k] >= 1) {
					cnt[k]--;
					break;
				}
			if (k > 'z') {
				puts("No");
				return 0;
			}
		}

	puts("Yes");

	return 0;
}