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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int h = sc.nextInt();
        int w = sc.nextInt();
        char[][] a = new char[h][w];
        int[] alpha = new int[26];
        for (int i = 0; i < h; i++) {
            a[i] = sc.next().toCharArray();
            for (int j = 0; j < w; j++) {
                alpha[a[i][j] - 'a']++;
            }
        }
        int g1;
        int g2;
        int g4;
        if (h % 2 == 0 && w % 2 == 0) {
            g1 = 0;
            g2 = 0;
            g4 = (h / 2) * (w / 2);
        } else if (h % 2 == 0 || w % 2 == 0) {
            g1 = 0;
            g2 = h / 2 + w / 2;
            g4 = (h / 2) * (w / 2);
        } else {
            g1 = 1;
            g2 = h / 2 + w / 2;
            g4 = (h / 2) * (w / 2);
        }
        for (int i = 0; i < alpha.length; i++) {
            if (g1 == 0) {
                break;
            }
            if (alpha[i] % 4 == 1 || alpha[i] % 4 == 3) {
                alpha[i]--;
                g1--;
                i = -1;
            }
        }
        for (int i = 0; i < alpha.length; i++) {
            if (g2 == 0) {
                break;
            }
            if (alpha[i] % 4 == 2) {
                alpha[i] -= 2;
                g2--;
                i = -1;
            }
        }
        for (int i = 0; i < alpha.length; i++) {
            if (g4 == 0) {
                break;
            }
            if (alpha[i] > 0 && alpha[i] % 4 == 0) {
                alpha[i] -= 4;
                g4--;
                i = -1;
            }
        }
        for (int i = 0; i < alpha.length; i++) {
            if (alpha[i] != 0) {
                System.out.println("No");
                return;
            }
        }
        System.out.println("Yes");
    }

}