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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

from collections import Counter

def solve():
    h, w = map(int, input().split())
    al = [input() for _ in range(h)]
    word_dict = Counter(list("".join(al)))

    g1, g2, g4 = 1, int(h/2) + int(w/2), int(h/2) * int(w/2)
    for _ in range(g1):
        key = 0
        for k, v in word_dict.most_common():
            if v % 4 == 1 or v % 4 == 3:
                key = k
                break
        if key is not 0:
            word_dict[key] -= 1
    
    for _ in range(g2):
        key = 0
        for k, v in word_dict.most_common():
            if v % 4 == 2:
                key = k
                break
        if key is not 0:
            word_dict[key] -= 2
    
    for _ in range(g4):
        key = 0
        for k, v in word_dict.most_common():
            if v % 4 == 0:
                key = k
                break
        if key is not 0:
            word_dict[key] -= 4
    
    print("Yes" if len([k for k, v in word_dict.most_common() if v != 0]) == 0 else "No")
    
    
if __name__=="__main__":
    solve()