Problem:
Score : 700 points

Problem Statement

There is an H \times W grid (H vertical, W horizontal), where each square contains a lowercase English letter.
Specifically, the letter in the square at the i-th row and j-th column is equal to the j-th character in the string S_i.

Snuke can apply the following operation to this grid any number of times:

Choose two different rows and swap them. Or, choose two different columns and swap them.

Snuke wants this grid to be symmetric.
That is, for any 1 \leq i \leq H and 1 \leq j \leq W, the letter in the square at the i-th row and j-th column and the letter in the square at the (H + 1 - i)-th row and (W + 1 - j)-th column should be equal.

Determine if Snuke can achieve this objective.

Constraints

1 \leq H \leq 12

1 \leq W \leq 12

|S_i| = W

S_i consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

H W
S_1
S_2
:
S_H

Output

If Snuke can make the grid symmetric, print YES; if he cannot, print NO.

Sample Input 1

2 3
arc
rac

Sample Output 1

YES

If the second and third columns from the left are swapped, the grid becomes symmetric, as shown in the image below:

Sample Input 2

3 7
atcoder
regular
contest

Sample Output 2

NO

Sample Input 3

12 12
bimonigaloaf
faurwlkbleht
dexwimqxzxbb
lxdgyoifcxid
ydxiliocfdgx
nfoabgilamoi
ibxbdqmzxxwe
pqirylfrcrnf
wtehfkllbura
yfrnpflcrirq
wvcclwgiubrk
lkbrwgwuiccv

Sample Output 3

YES

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main(){
  int h,w,i,j,k,l,c=0,d=0,b[2][20][30]={};
  int f[2][30]={};
  char s[20][20];
  scanf("%d %d",&h,&w);
  for(i=0;i<h;i++)scanf("%s",s[i]);
  for(i=0;i<h;i++){
    for(j=0;j<w;j++)b[0][i][s[i][j]-'a']++;
    //for(j=0;j<26;j++)printf("%c%d ",j+'a',b[0][i][j]);printf("\n");
  }//printf("\n");
  for(i=0;i<w;i++){
    for(j=0;j<h;j++)b[1][i][s[j][i]-'a']++;
    //for(j=0;j<26;j++)printf("%c%d ",j+'a',b[1][i][j]);printf("\n");
  }//printf("\n");
  for(l=0;l<2;l++){
    for(i=0;i<(l?w:h);i++){
      if(f[l][i])continue;
      for(j=i+1;j<(l?w:h);j++){
	if(f[l][j])continue;
	for(k=0;k<26;k++){
	  if(b[l][i][k]-b[l][j][k])break;
	}
	if(k==26)f[l][i]=f[l][j]=i+1;
      }
    }
  }
  //for(i=0;i<h;i++)printf("%d\n",f[0][i]);
  //for(i=0;i<w;i++)printf("%d\n",f[1][i]);
  for(i=0;i<h;i++){
    if(f[0][i]==0)c++;
  }
  for(i=0;i<w;i++){
    if(f[1][i]==0)d++;
  }
  printf("%s\n",c==h%2&&d==w%2?"YES":"NO");
  return 0;
}