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>
#include <stdlib.h>
int comp(const void *a, const void *b) {
    return *(char*)a - *(char*)b;
}
int main() {
  int h,w,i,j,k;
  char s[12][13],a[145]={0};
  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++) {
      a[k++]=s[i][j];
    }
  }
  qsort(a, h*w, sizeof(char), comp);
  if ((h*w)%2==0) {
    for (i=0;i<h*w;i+=2) {
      if (a[i]!=a[i+1]) {
        printf("NO\n");
        return 0;
      }
    }
  } else {
    j=0;
    for (i=0; i<h*w; i+=2) {
      if (a[i]!=a[i+1]) {
        j+=1;
        i-=1;
        if (j>1) {
          printf("NO\n");
          return 0;
        }
      }
    }
  }
  printf("YES\n");
  return 0;
}