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: Wrong Answer)

H, W = [int(x) for x in input().split(" ")]

d = {}

for h in range(H):
    row = input()
    for a in list(row):
        if a in d:
            d[a] += 1
        else:
            d[a] = 1

print(d)

res = False

if H % 2 == 0 and W % 2 == 0:
    for v in d.values():
        if v % 4 != 0:
            res = True
            break
    res = not res

elif H % 2 == 1 and W % 2 == 0:
    cnt4 = (W // 2) * ((H - 1) // 2)
    cnt2 = (W // 2)
    temp4 = 0
    temp2 = 0
    f = True
    for v in d.values():
        if v % 2 != 0:
            f = False
            break

        temp4 += v // 4
        temp2 += 1 if v % 4 != 0 else 0
    
    if f and cnt2 >= temp2 and (cnt2 - temp2) % 2 == 0:
        res = True
elif H % 2 == 0 and W % 2 == 1:
    cnt4 = (H // 2) * ((W - 1) // 2)
    cnt2 = (H // 2)
    temp4 = 0
    temp2 = 0
    f = True
    for v in d.values():
        if v % 2 != 0:
            f = False
            break

        temp4 += v // 4
        temp2 += 1 if v % 4 != 0 else 0
    
    if f and cnt2 >= temp2 and (cnt2 - temp2) % 2 == 0:
        res = True
else:
    cnt4 = ((H - 1) // 2) * ((W - 1) // 2)
    cnt2 = ((H - 1) // 2) + ((W - 1) // 2)
    temp4 = 0
    temp2 = 0
    temp1 = 0

    for v in d.values():
        temp4 += v // 4
        temp2 += (v % 4) // 2
        temp1 += (v % 2)
    
    if temp1 == 1 and cnt2 >= temp2 and (cnt2 - temp2) % 2 == 0:
        res = True

print("Yes" if res else "No")