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 = list(map(int , input().split()))
D = []
for j in range(H):
    x=list(input())
    D.extend(x)
s = len((set(D)))
c = collections.Counter(D)
Q = []

for i in range(s):
    Q.append(c.most_common()[i][1])
if s == 1:
    print("Yes")
else:
    if ((H*W)%2 == 1):
        seigen = (H//2+1)*(W//2 +1)
        q = 0
        for k in range(s):
            q += (Q[k] % 2)
        if (q == 1)and(s<=seigen):
            print("Yes")
        else:
            print("No")
    elif ((H%2) == 0)and((W %2) ==0):
        q = 0
        for k in range(s):
            q += (Q[k] % 4)
        if ((s % 2) == 0)and(q == 0):
            print("Yes")
        else:
            print("No")
    
    else:
        q = 0
            
        if ((H%2) == 1):
            seigen = (H//2+1)*W
        elif ((W%2) == 1):
            seigen = (W//2+1)*H
        for k in range(s):
            q += (Q[k] % 2)
        if (q == 0)and(s<=seigen):
            print("Yes")
        else:
            print("No")