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: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int h = sc.nextInt();
        int w = sc.nextInt();

        Map<Character,Integer> map = new HashMap<>();
        StringBuffer sb = new StringBuffer();
        for(int i=0 ;i<h;i++) {
            sb.append(sc.next());
        }
//        System.out.println(sb);

        for(int i=0 ;i<h*w;i++){
            char c  =  sb.charAt(i);
            int count = map.getOrDefault(c,0);
            map.put(c,count+1);
        }


        int need4 = (h/2)*(w/2);
        int need1 = 0;
        if((h%2 == 1) && (w%2 == 1)){
            need1=1;
        }

        int need2 = 0;
        if(h%2==1){
            need2+=w/2;
        }

        if(w%2==1){
            need2+=h/2;
        }

        int map4 = 0;
        int map2 = 0;
        int map1 = 0;

        Set<Character> charUniq = map.keySet();

        for(Character c : charUniq){
            int count = map.get(c);
            map4 += count/4;
            count = count%4;
            map.put(c,count);

        }

        for(Character c : charUniq){
            int count = map.get(c);
            map2 += count/2;
            count = count%2;
            map.put(c,count);
        }

        for(Character c : charUniq){
            int count = map.get(c);
            if(1 == count){
                map1 += 1;
            }
        }
//
//        System.out.println(map1);
//        System.out.println(map2);
//        System.out.println(map4);
//
//        System.out.println(need1);
//        System.out.println(need2);
//        System.out.println(need4);


        //////

        map4 = map4 - need4;
        if(map4 < 0 ){
            System.out.print("No");
            return;
        }

        map2 += map4 * 2;
        map2 = map2 - need2;
        if(map2 < 0){
            System.out.print("No");
            return;
        }

        if(map2 == 0 && map1==need1){
            System.out.print("Yes");
            return;
        }
        System.out.print("No");

    }
}