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: Accepted)

# coding: utf-8
h,w=map(int,input().split())
dic={}
for i in range(h):
    for c in input():
        if c in dic:
            dic[c]=(dic[c]+1)%4
        else:
            dic[c]=1
d=list(dic.values())
if h%2 and w%2:
    if d.count(1)+d.count(3)==1 and d.count(2)<=h//2+w//2:
        print('Yes')
    else:
        print('No')
elif (h%2 and not w%2) or (w%2 and not h%2):
    if not (d.count(1)+d.count(3)>0):
        if h%2:
            if d.count(2)<=w//2 :#and d.count(2)%2==(w//2)%2:
                print('Yes')
            else:
                print('No')
        elif w%2:
            if d.count(2)<=h//2 :#and d.count(2)%2==(h//2)%2:
                print('Yes')
            else:
                print('No')
    else:
        print('No')
else:
    if sum(d)==0:
        print('Yes')
    else:
        print('No')