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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

import Control.Applicative
import Data.List

main = do
  [h,w] <- map read . words <$> getLine :: IO [Int]
  s <- getContents

  let count = [length $ filter (==c) s | c <- ['a'..'z']]
  
  putStrLn $ if solve h w count then "Yes" else "No"
  
  where
    solve h w count
      | even h && even w = all ((==0) . (`mod`4)) count
      | odd h && odd w =
        (length . filter ((==2) . (`mod`4))) count <= (h `div` 2 + w `div` 2) &&
        (length . filter odd) count == 1
      | odd h = all even count && (length . filter ((==2) . (`mod`4))) count <= w `div` 2
      | odd w = all even count && (length . filter ((==2) . (`mod`4))) count <= h `div` 2