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: Wrong Answer)

{-# LANGUAGE FlexibleContexts #-}
import Data.Array
import Data.Array.ST

type CountVect=(Int,Int,Int,Int)

case_ee::CountVect->(Int,Int)->Bool
case_ee (a,b,c,d) _     = (b,c,d)==(0,0,0)

case_eo::CountVect->(Int,Int)->Bool
case_eo (a,b,c,d) (e,o) = let limit = div e 2 in ((b,d)==(0,0)) && (c <= limit)

case_oo::CountVect->(Int,Int)->Bool
case_oo (a,b,c,d) (h,w) = case (b,d) of
                          (1,0) -> (b <= (div (h+w-2) 2)  )
                          (0,1) -> (b <= (div (h+w-2) 2)-1)
                          (0,0) -> False
                          otherwise -> False

preproc1::String->Array Char Int
preproc1 str = runSTArray $ do
                  ar<-newArray ('a','z') 0
                  mapM_ (count ar) str
                  return ar
                        where count ar x = do
                                v<-(readArray ar x)
                                writeArray ar x (v+1)

preproc2::Array Char Int->CountVect
preproc2 ar = let ls = elems ar in
              let  a = length (filter (\x->mod x 4 == 0) ls) in
              let  b = length (filter (\x->mod x 4 == 1) ls) in
              let  c = length (filter (\x->mod x 4 == 2) ls) in
              let  d = length (filter (\x->mod x 4 == 3) ls) in
              (a,b,c,d)

solver::String->(Int,Int)->Bool
solver str (h,w)=let vec = (preproc2 (preproc1 str)) in
                  case (mod h 2,mod w 2) of
                  (0,0) -> case_ee vec (h,w)
                  (1,0) -> case_eo vec (w,h)
                  (0,1) -> case_eo vec (h,w)
                  (1,1) -> case_ee vec (h,w)


main::IO()
main=do
  size<-getLine
  let h:w:[]= map read (words size)
  str<-getdat h
  putStr (if (solver str (h,w)) then "Yes" else "No")
  putStr "\n"

getdat::Int->IO String
getdat 0 = return []
getdat n = do
  str<-getLine
  rest<-getdat (n-1)
  return (str++rest)