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)

import collections

H, W = map(int, input().split())
A = []
for _ in range(H):
    A += list(input())
elements = collections.Counter(A)
c = collections.Counter([x % 4 for x in elements.values()])
keys = sorted(list(c.keys()))

ans = False
if H % 2 == 0 and W % 2 == 0:
    if keys == [0]:
        ans = True

elif H % 2 != 0 and W % 2 != 0:
    if (1 in c.keys() and c[1] == 1) ^ (3 in c.keys() and c[3] == 1):
        if 1 in c.keys():
            c[0] += 1
            keys.pop(keys.index(1))
            c.pop(1)
        else:
            c[2] += 1
            keys.pop(keys.index(3))
            c.pop(3)
        
        if keys == [0] or c[2] <= (H + W)//2 - 1:
            ans = True 

elif H % 2 != 0:
    if keys == [0] or ((keys == [0,2] or keys == [2]) and c[2] <= W//2):
        ans = True
else:
    if keys == [0] or ((keys == [0,2] or keys == [2]) and c[2] <= H//2):
        ans = True

if ans:
    print('Yes')
else:
    print('No')