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)

def fourcount(List):

	num = 0
	tmp = 1
	l = len(List)

	for i in range(1,l):

		if (List[i-1] == List[i]) and (tmp < 4):
			tmp += 1

		else:
			if tmp == 4:
				num += 1
			
			tmp = 1 
		
	return num	

def twocount(List):

	num = 0
	tmp = 1
	l = len(List)

	for i in range(1,l):

		if (List[i-1] == List[i]) and (tmp < 2):
			tmp += 1

		else:
			if tmp == 2:
				num += 1
			
			tmp = 1 
		
	return num

def decision(H,W,num4,num2):

	if H == 1:
		if (W % 2==0) and (num2 == W/2):
			print("Yes")
		elif (W % 2 == 1) and (num2 == (W-1)/2):
			print("Yes")
		else:
			print("No")

	elif W == 1:
		if (H % 2==0) and (num2 == H/2):
			print("Yes")
		elif (H % 2 == 1) and (num2 == (H-1)/2):
			print("Yes")
		else:
			print("No")

	elif H > W:
		if (H % 2 == 0) and (H / 2 <= num4):
			same = H/2
			decision(H,W-2,num4-same,num2)
		elif (H % 2 == 1) and ((H-1) / 2 <= num4 ) and (num2 >= 1):
			same =  (H-1) /2
			decision(H,W-2,num4-same,num2-1)
		else:
			print("No")

	elif W > H:
		if (W % 2 == 0) and (W / 2 <= num4):
			same = H/2
			decision(H-2,W,num4-same,num2)
		elif (W % 2 == 1) and ((W-1) / 2 <= num4) and (num2 >= 1):
			same =  (W-1) /2
			decision(H-2,W,num4-same,num2-1)
		else:
			print("No")

	elif W == H:
		if (H % 2 == 0) and (H - 1) <= num4 :
			decision(H-2,W-2,num4-(H-1),num2)
		elif (H % 2 == 1) and (H - 2) <= num4 and (num2 >= 2):
			decision(H-2,W-2,num4-(H-2),num2-2)
		else :
			print("No")



H,W = map(int,input().split(" "))
a = []
b = []

for i in range(H):
	a = list(input())
	a = a[0:W]
	for string in a:
		b.append(string)

b.sort()

num4 = fourcount(b)
num2 = twocount(b) - (num4*2)

decision(H,W,num4,num2)