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 Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

program name
    implicit none
    integer(4):: h,w,cnt
    integer(4):: a_cnt(ichar('a'):ichar('z'))
    character(:),allocatable:: a
    integer(4):: i,j
    logical :: b

    read*, h,w
    allocate(character(w):: a)
    a_cnt(:) = 0
    do i=1,h
        read*, a(:)
        do j=1,w
            a_cnt(ichar(a(j:j))) = a_cnt(ichar(a(j:j))) + 1
        end do
    end do

    a_cnt(:) = mod(a_cnt(:),4)
    
    if (mod(w,2) == 0 .and. mod(h,2) == 0) then
        b = sum(a_cnt) == 0
    else if(mod(w,2) /= 0 .and. mod(h,2) /= 0) then
        b = sum(mod(a_cnt,2)) <= 1
    else
        b = sum(mod(a_cnt,2)) == 0
    end if

    if (b) then
        print*, 'Yes'
    else
        print*, 'No'
    end if
end program name