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 Data.List

dfs :: Int -> Int -> [Int] -> [Int] -> (Bool, [Int])
dfs gn r cs ds
  | r < 1 = (True, ds ++ cs)
  | null cs = (False, ds)
  | otherwise = dfs gn r' cs' ds'
  where
    c = head cs
    r' = r - (c `div` gn)
    cs' = tail cs
    ds' = ds ++ [c `mod` gn]

solve (l1:l2)
  | result' && sum cs'' == 0 = "Yes"
  | otherwise = "No"
  where
    (h:w:_) = map read $ words l1
    cs = map length $ group $ sort $ concat l2
    g4 = (h `div` 2) * (w `div` 2)
    g2
      | odd h && odd w = (h `div` 2) + (w `div` 2)
      | odd h && even w = w `div` 2
      | even h && odd w = h `div` 2
      | otherwise = 0
    g1 = if odd h && odd w then 1 else 0
    step (result, cs') (gn, r)
      | result = dfs gn r cs' []
      | otherwise = (False, [])
    (result', cs'') = foldl' step (dfs 4 g4 cs []) [(2, g2), (1, g1)]

main = getContents >>= putStrLn.solve.lines