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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int h = sc.nextInt();
		int w = sc.nextInt();

		int[] alphabet = new int[26];
		for (int i = 0; i < alphabet.length; i++) {
			alphabet[i] = 0;
		}

		for (int i = 0; i < h; i++) {
			String s = sc.next();
			for (int j = 0; j < w; j++) {
				char c = s.charAt(j);
				int ascii = (int) c;
				int alphabetIdx = ascii - 97;
				alphabet[alphabetIdx]++;
			}
		}

		int x1 = 0;
		int x2 = 0;
		int x4 = 0;
		for (int i = 0; i < alphabet.length; i++) {
			if (alphabet[i] % 4 == 0) {
				x4++;
			} else if (alphabet[i] % 2 == 0) {
				x2++;
			} else {
				x1++;
			}
		}

		if (h % 2 == 0 & w % 2 == 0) {
			if (x4 == h * w / 4) {
				System.out.print("Yes");
			} else {
				System.out.print("No");
			}
		} else if (h % 2 != 0 & w % 2 != 0) {
			if (x1 == 1 && x2 == (h - 1) / 2 + (w - 1) / 2) {
				System.out.print("Yes");
			} else {
				System.out.print("No");
			}
		} else if (h % 2 == 0 & w % 2 != 0) {
			if (x1 == 0 && x2 == h / 2) {
				System.out.print("Yes");
			} else {
				System.out.print("No");
			}
		} else if (h % 2 != 0 & w % 2 == 0) {
			if (x1 == 0 && x2 == w / 2) {
				System.out.print("Yes");
			} else {
				System.out.print("No");
			}
		}
	}
}