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.io.*;
import java.util.*;

public class Main {
	public static boolean enoughTwos(int r, int amount) {
		return (r <= amount/2) && ((r % 2) == (amount % 2));
	}
	public static void main(String[] args) throws IOException {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(f.readLine());
		int h = Integer.parseInt(st.nextToken());
		int w = Integer.parseInt(st.nextToken());
		int counts[] = new int[26];
		for(int i = 0; i < h; i++) {
			String s = f.readLine();
			for(int j = 0; j < w; j++) {
				counts[s.charAt(j) - 'a']++;
			}
		}
		int residues[] = new int[4];
		for(int i = 0; i < 26; i++) {
			residues[counts[i] % 4]++;
		}
		boolean good = true;
		int oddCount = 0;
		if(h % 2 == 1) {
			oddCount++;
		}
		if(w % 2 == 1) {
			oddCount++;
		}
		if(oddCount == 0) {
			good = (residues[1] == 0) && (residues[2] == 0) && (residues[3] == 0);
		} else if(oddCount == 1) {
			if(h % 2 == 1) {
				good = (residues[1] == 0) && enoughTwos(residues[2], w) && (residues[3] == 0);
			} else {
				good = (residues[1] == 0) && enoughTwos(residues[2], h) && (residues[3] == 0);
			}
		} else if(oddCount == 2) {
			good = (residues[1] == 1) && enoughTwos(residues[2], h - 1 + w - 1) && (residues[3] == 0);
		}
		if(good) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}
	}
}