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`.

Examples

Input

2 3
arc
rac


Output

YES


Input

3 7
atcoder
regular
contest


Output

NO


Input

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


Output

YES

#include <bits/stdc++.h>

using namespace std;

char a[15][15], pc[15], pr[15];
int n, m;
inline bool mp_r(int x);
inline bool mp_c(int x);

bool mp_c(int x) {
  if (x>m) 
    return mp_r(1);
  if (pc[x])
    return mp_c(x+1);

  int t = 0;
  for (int i = x; i <= m; ++i) 
    if (!pc[i]) ++t;
  for (int i = x; i <= m; ++i) {
    if (pc[i]) continue;
    pc[x] = i; pc[i] = x;
    if (((t & 1) || (i != x)) && mp_c(x+1))
      return true;
    pc[x] = 0; pc[i] = 0;
  }

  return false;
}

bool mp_r(int x) {
  if (x>n) 
    return true;
  if (pr[x])
    return mp_r(x+1);

  int t = 0;
  for (int i = x; i <= n; ++i)
    if (!pr[i]) ++t;
  for (int i = x; i <= n; ++i) {
    if (pr[i]) continue;
    bool ok = true;
    for (int j = 1; j <= m; ++j) 
      ok &= (a[x][j] == a[i][pc[j]]);
    if (((t & 1) || (i != x)) && ok) {
      pr[x] = i; pr[i] = x;
      if (mp_r(x+1)) return true;
      pr[x] = 0; pr[i] = 0;
    }
  }
  
  return false;
}

int main() {
  cin >> n >> m;
  for(int i=1;i<=n;++i) scanf("%s", a[i]+1);

  if (mp_c(1)) {
    puts("YES");
  } else {
    puts("NO");
  }

  return 0;
}
