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;
import java.util.stream.Collectors;

class Main {

    public static void main(String[] args) {
        new Main().compute();
    }

    void compute() {
        Scanner sc = new Scanner(System.in);
        int H = sc.nextInt();
        int W = sc.nextInt();
        int cap_1 = H % 2 == 1 && W % 2 == 1 ? 1 : 0;
        int cap_2 = H % 2 + W % 2;
        int[] appearance = new int[26];
        for (int i = 0; i < H; i++) {
            String a = sc.next();
            for (int j = 0; j < W; j++) {
                appearance[a.charAt(j) - 'a']++;
            }
        }
        for (int i = 0; i < 26; i++) {
            if (appearance[i] % 2 == 1) {
                cap_1--;
            } else if (appearance[1] % 4 == 2) {
                cap_2--;
            }
            if (cap_1 < 0 || cap_2 < 0) {
                System.out.println("No");
                return;
            }
        }
        System.out.println("Yes");
    }
}