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 C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int main(){
  int n,m,a=0,c=0,d,i,j,b[30]={};
  char s;
  scanf("%d %d",&n,&m);
  for(i=0;i<n;i++){
    for(j=0;j<m;j++){
      scanf(" %c",&s);
      b[s-'a']++;
    }
  }
  for(i=0;i<26;i++){
    if(b[i]%4==0)continue;
    if(b[i]%4==2)a+=2;
    if(b[i]%2){
      if(b[i]%4==3)a+=2;
      c++;
    }
  }
  //printf("%d %d\n",(n%2)*(m%2),c);
  //printf("%d %d\n",n/2*2*(m%2)+m/2*2*(n%2),a);
  d=n/2*2*(m%2)+m/2*2*(n%2);//printf("%d %d\n",d,a);
  if(c-((n%2)*(m%2)))printf("No\n");
  else if(d<a)printf("No\n");
  else{
    if(d%4==a%4)printf("Yes\n");
    else printf("No\n");
  }
  //else printf("Yes\n");
  return 0;
}