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.*;
 
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int H = sc.nextInt();
        int W = sc.nextInt();
        int[] num = new int[26];
        for (int i = 0; i < H; i++) {
            String s = sc.next();
            for (int j = 0; j < W; j++) {
                num[s.charAt(j)-'a']++;
            }
        }

        boolean ans = true;
        if (H%2 == 0 && W%2 == 0) {
            int num4 = H*W/4;
            for (int i = 0; i < 26; i++) {
                if (num[i] >= 4) {
                    int a = Math.min(num[i]/4, num4);
                    num[i] -= a*4;
                    num4 -= a;
                }
            }
            int others = 0;
            for (int i = 0; i < 26; i++) {
                others += num[i];
            }
            ans = others == 0;
        } else if (H%2 == 1 && W%2 == 1) {
            int num4 = (H-1)*(W-1)/4;
            for (int i = 0; i < 26; i++) {
                if (num[i] >= 4) {
                    int a = Math.min(num[i]/4, num4);
                    num[i] -= a*4;
                    num4 -= a;
                }
            }
            int num2 = ((H-1)+(W-1))/2;
            for (int i = 0; i < 26; i++) {
                if (num[i] >= 2) {
                    int a = Math.min(num[i]/2, num2);
                    num[i] -= a*2;
                    num2 -= a;
                }
            }
            int others = 0;
            for (int i = 0; i < 26; i++) {
                others += num[i];
            }
            ans = others == 1;
        } else {
            int odd = H%2 == 1 ? H : W;
            int even = H%2 == 1 ? W : W;

            int num4 = (odd-1)*even/4;
            for (int i = 0; i < 26; i++) {
                if (num[i] >= 4) {
                    int a = Math.min(num[i]/4, num4);
                    num[i] -= a*4;
                    num4 -= a;
                }
            }
            int num2 = even/2;
            for (int i = 0; i < 26; i++) {
                if (num[i] >= 2) {
                    int a = Math.min(num[i]/2, num2);
                    num[i] -= a*2;
                    num2 -= a;
                }
            }
            int others = 0;
            for (int i = 0; i < 26; i++) {
                others += num[i];
            }
            ans = others == 0;
        }
        System.out.println(ans ? "Yes" : "No");
    }
}