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 Python (3.4.3) (judge verdict: Wrong Answer)

# -*- coding: utf-8 -*-


def main():
    h, w = map(int, input().split())
    alpha = [0 for _ in range(26)]

    # 各アルファベットの個数を数える
    for i in range(h):
        ai = list(input())

        for aii in ai:
            # アルファベットを数値に変換
            index = ord(aii) - ord('a')
            alpha[index] += 1

    # 回文となる条件
    # 1. hとwがともに偶数：
    # アルファベットの個数が全て4の倍数のとき，Yes
    # 2. hとwのどちらかが偶数：
    # アルファベットの個数が全て2の倍数のとき，Yes
    # 3. hとwがともに奇数：
    # (h - 1) * (w - 1)マス分，アルファベットの個数が4の倍数であり，
    # かつ，1種類を除いて，2の倍数のとき，Yes
    if h % 2 == 0 and w % 2 == 0:
        for a in alpha:
            if a % 4 != 0:
                print('No')
                exit()

        print('Yes')
    elif h % 2 == 0 or w % 2 == 0:
        for a in alpha:
            if a % 2 != 0:
                print('No')
                exit()

        print('Yes')
    else:
        need = (h - 1) * (w - 1)

        for index, a in enumerate(alpha):
            p, q = divmod(a, 4)
            alpha[index] -= min(p * 4, need)
            need -= min(p * 4, need)

            if need == 0:
                break

        count = 0

        for a in alpha:
            if a % 2 == 1:
                count += 1

        if count == 1:
            print('Yes')
        else:
            print('No')


if __name__ == '__main__':
    main()