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)

t="abcdefghijklmnopqrstuvwxyz"
h,w=map(int,input().split())
a=[list(input())for _ in range(h)]
b=[0]*26
for i in a:
  for j in i:b[t.index(j)]+=1
x,y,z=(h//2)*(w//2),(h%2)*(w//2)+(h//2)*(w%2),(h%2)*(w%2)
for i in b:
  j=i
  while x and j>3:
    x-=1
    j-=4
  while y and j>1:
    y-=1
    j-=2
  z-=j
if not(x==y==z==0):print("No")
else:print("Yes")
"""#一例出力、問題誤読して実装した
c=[w*[0]for _ in range(h)]
fi=ti=0
for i in range(27):
  while b[i]:
    if b[i]>=4:
      j,k=fi//(w//2),fi%(w//2)
      c[j][k]=c[j][w-k-1]=c[h-j-1][k]=c[h-j-1][w-k-1]=t[i]
      fi+=1
      b[i]-=4
    if b[i]==2:
      if ti<h//2 and w%2:c[ti][w//2]=c[h-ti-1][w//2]=t[i]
      else:c[h//2][ti]=c[h//2][w-1-ti]=t[i]
      ti+=1
      b[i]-=2
    if b[i]==1:
      c[h//2][w//2]=t[i]
      b[i]-=1
for i in c:print("".join(i))
"""