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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import collections
h,w = map(int,input().split())
ab = []
for _ in range(h):
    s=list(input())
    c = collections.Counter(s)
    c = sorted(c.items(), key=lambda x:-x[1])
    kisu_count = 0
    for i in c:
        if i[1]%2==1:
            kisu_count += 1
    if w%2==1 and kisu_count == 1 or w%2==0 and kisu_count == 0:
        s.sort()
        s=''.join(map(str,s))
        ab.append(s)
    else:
        print('No')
        exit()

memo = []

count = 1

ab.sort()

for i in range(1,h):
    if ab[i] == ab[i-1]:
        count += 1
    else:
        memo.append(count%2)
        count = 1

memo.append(count%2)

kisu_count2 = 0

for i in memo:
    if i == 1:
        kisu_count2 += 1

if h%2==1 and kisu_count2 == 1 or h%2==0 and kisu_count2 == 0:
    print('Yes')
else:
    print('No')