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 (2.7.6) (judge verdict: Wrong Answer)

h, w = map(int, raw_input().split())
d = {}
for i in range(h):
    t = raw_input()
    for j in range(w):
        if d.has_key(t[j]):
            d[t[j]] += 1
        else:
            d[t[j]] = 1

two = 0
one = 0
four = 0
if w < 2:
    two = h // 2
    one = h % 2
elif h < 2:
    two = w //2
    one = h % 2
elif h%2 == 0:
    if w%2 == 0:
        four = h*w/4
    else:
        four = h//2 + (w-2)//2
        two = h//2
else:
    if w%2 == 0:
        four = w//2 + (h-2)//2
        two = w//2
    else:
        four = w//2 + (h-2)//2
        one = 1
        two = w//2 + h//2

def calc(dic, num, n):
    for key, value in dic.items():
        if n==0:
            return True, dic
        if value >= num:
            dic[key] -= num
            n -= 1
    if n:
        return False, dic
    else:
        return True, dic

result, d = calc(d, 4, four)
if result:
    result, d = calc(d, 2, two)
if result:
    result, d = calc(d, 1, one)

if result:
    print 'Yes'
else:
    print 'No'