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 (3.8.2) (judge verdict: Wrong Answer)

from collections import Counter,defaultdict,deque
from heapq import heappop,heappush
from bisect import bisect_left,bisect_right 
import sys,math,itertools,fractions,pprint
sys.setrecursionlimit(10**8)
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))

h,w = inpl()
cnt = defaultdict(int)
for _ in range(h):
    for x in input():
        cnt[x] += 1
one = two = 0
for va in cnt.values():
    now = va%4
    if now == 1 or now == 3:
        one += 1
    if now == 2 or now == 3:
        two += 1
res = False
if h%2 == 0 and w%2 == 0:
    if not one and not two:
        res = True
elif h%2 and w%2:
    if one == 1 and two <= (w-1)//2 + (h-1)//2:
        res = True
else:
    now = h if w%2 else w
    if not one and two <= w//2:
        res = True
print('Yes' if res else 'No')