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 {
	static Scanner sc = new Scanner(System.in);

	public static void main(String[] args) throws Exception {
		int n=sc.nextInt();
		int m=sc.nextInt();
		sc.nextLine();
		int[] map=new int[128];
		for(int i=0;i<n;i++){
			String str=sc.nextLine();
			for(char c:str.toCharArray()){
				map[c]++;
			}
		}

		int one=0;
		int two=0;
		int four=0;
		boolean nTag=((n&1)==1);
		boolean mTag=((m&1)==1);
		if(nTag&&mTag){
			one=1;
			two=n+m-2;
			four=n*m-one-two;
		}
		else if(nTag){
			two=m;
		}
		else if(mTag){
			two=n;
		}
		four=n*m-one-two;	
	
		for(int i='a';i<='z';i++){
			int res=map[i]/4*4;
			four-=res;
			two-=(map[i]%4)/2*2;
			map[i]-=two;
			one-=(map[i]&1);
		}

		if(one==0&&two==0&&four==0)
			System.out.println("Yes");
		else
			System.out.println("No");		
		
	}
}