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 check(int,int,int*);

int main(){
  int H,W;
  int answer;
  scanf("%d%*c%d\n",&H,&W);
  char a;
  int array[26];
  for(int i=0;i<26;i++){
    array[i] = 0;
  }
  int k;
  for(int i=0;i<H;i++){
    for(int j=0;j<W-1;j++){
      scanf("%c",&a);
      k = a - 'a';
      array[k] += 1;
    }
    scanf("%c\n",&a);
    k = a - 'a';
    array[k] += 1;
  }
  int amari[4];
  for(int i=0;i<4;i++){
    amari[i]=0;
  }
  for(int i=0;i<26;i++){
    amari[0] += array[i]/4;
    k = array[i]%4;
    if(k != 0){
      amari[k]++;
    }
  }
  if(amari[1]+amari[3] > 1){
    answer = 0;
  }else{
    answer = check(H,W,amari);
  }
  if(answer == 1){
    printf("Yes");
  }else{
    printf("No");
  }
  return 0;
}
 
int check(int H,int W,int *amari){
  if(H>1 && W>1){
    if(amari[0] >= (H/2)*(W/2)){
      amari[0] -= (H/2)*(W/2);
      if(H%2 == 0 && W%2 == 0){
	return 1;
      }
    }else{
      return 0;
    }
  }
  amari[2] += amari[3] + amari[0]*2;
  if(H%2 == 1){
    amari[2] -= W/2;
  }
  if(W%2 == 1){
    amari[2] -= H/2;
  }
  if(amari[2] == 0){
    return 1;
  }else{
    return 0;
  }
}