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)

from collections import Counter
def solve():
    inputs = open(0).readlines()
    h, w = map(int, inputs[0].split())
    *A, = map(lambda x: x.strip(), inputs[1:])
    B = []
    for S in A:
        C = [0] * 26
        for c in S:
            C[ord(c)-ord('a')] += 1
        B.append(C)
    
    if w % 2 == 0:
        if all(all(x % 2 == 0 for x in b) for b in B):
            C = []
        else:
            return 'No'
    else:
        C = [[i for i, x in enumerate(b) if x % 2 != 0] for b in B]
        if all(len(c) == 1 for c in C):
            B = [[x if x % 2 == 0 else x-1 for x in b] for b in B]
            C = Counter([c[0] for c in C]).values()
        else:
            return 'No'
    C
    D = Counter(map(tuple, B)).values()
    if h % 2 == 0:
        if all(c % 2 == 0 for c in C) and all(d % 2 == 0 for d in D):
            return 'Yes'
    else:
        if sum(c % 2 for c in C) <= 1 and sum(d % 2 for d in D) <= 1:
            return 'Yes'
    return 'No'

print(solve())