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 Python (3.4.3) (judge verdict: Accepted)

from collections import Counter
H, W = map(int, input().split())
cnt4, cnt2, cnt1 = 0, 0, 0
for v in Counter(''.join(input() for _ in range(H))).values():
    if v % 4 == 0:
        continue
    elif v % 2 == 0:
        cnt2 += 1
    else:
        cnt1 += 1
        if (v - 1) % 4 != 0:
            cnt2 += 1
odd = (H % 2 == 1) + (W % 2 == 1)
if odd == 2:
    if cnt1 <= 1 and cnt2 <= (H + W - 2) // 2:
        print("Yes")
    else:
        print("No")
elif odd == 1 and H % 2 == 1:
    if cnt1 == 0 and cnt2 <= W // 2:
        print("Yes")
    else:
        print("No")
elif odd == 1 and W % 2 == 1:
    if cnt1 == 0 and cnt2 <= H // 2:
        print("Yes")
    else:
        print("No")
else:
    if cnt1 == 0 and cnt2 == 0:
        print("Yes")
    else:
        print("No")