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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int,int> P;

int W,H;
char F[15][15];

bool eq(int a,int b){
  vector<char> c1,c2;
  for(int i=0;i<W;i++){
    c1.emplace_back( F[a][i] );
    c2.emplace_back( F[b][i] );
  }
  sort( c1.begin(), c1.end() );
  sort( c2.begin(), c2.end() );
  for(int i=0;i<W;i++)
    if( c1[i] != c2[i] ) return false;
  return true;
}

void rotate(){
  char tmp[15][15];
  for(int i=0;i<H;i++)
    for(int j=0;j<W;j++)
      tmp[W-j-1][i] = F[i][j];
  swap(H,W);
  for(int i=0;i<H;i++)
    for(int j=0;j<W;j++)
      F[i][j] = tmp[i][j];
}

int hcheck(){
  bool f[15] = {};
  for(int i=1;i<H;i++){
    if( f[i] ) continue;
    for(int j=0;j<i;j++){
      if( f[j] ) continue;
      if( eq(i,j) ) {
        f[i] = f[j] = true; break;
      }
    }
  }
  int cnt = 0;
  for(int i=0;i<H;i++) {
    if( !f[i] ) cnt++;
  }
  if(!(H&1)&&cnt!=0) return false;
  if((H&1)&&cnt!=1) return false;
  return true;
}

bool solve(){
  if(!hcheck()) return false;
  rotate();
  return hcheck();  
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> H >> W;
  for(int i=0;i<H;i++)
    for(int j=0;j<W;j++)
      cin >> F[i][j];
  if( solve() ) cout << "YES" << endl;
  else cout << "NO" << endl;
}