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.Arrays;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		new Main().solve();
	}
	
	void solve(){
		Scanner sc=new Scanner(System.in);
		int H=sc.nextInt();
		int W=sc.nextInt();
		char[][]a=new char[H][W];
		String ans="Yes";
		for(int i=0;i<H;i++) {
			String s=sc.next();
			a[i]=s.toCharArray();
		}
		int[]alph=new int[26];
		for(int i=0;i<H;i++) {
			for(int j=0;j<W;j++) {
				alph[a[i][j]-'a']++;
			}
		}
		int g1=1;
		int g2=H/2*H%2+W/2*W%2;
		int g3=H/2*W/2;
		
		LOOP1:while(g1>0) {
			for(int i=0;i<26;i++) {
				if(alph[i]%4==1 || alph[i]%4==3){
					alph[i]-=1;
					g1--;
					continue LOOP1;
				}
			}
			break;
		}
		LOOP2:while(g2>0) {
			for(int i=0;i<26;i++) {
				if(alph[i]%4==2){
					alph[i]-=2;
					g2--;
					continue LOOP2;
				}
			}
			break;
		}
		LOOP3:while(g3>0) {
			for(int i=0;i<26;i++) {
				if(alph[i]>0 && alph[i]%4==0){
					alph[i]-=4;
					g3--;
					continue LOOP3;
				}
			}
			break;
		}
		for(int i=0;i<26;i++){
			if(alph[i]>0) {
				ans="No";
				break;
			}
		}
		System.out.println(ans);
	}
}