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;

int h, w;
mt19937 ei(1333333);

bool check(vector<string> s) {
  int h = s.size();
  vector<string> t = s;
  for (auto &i : t) reverse(i.begin(), i.end());

  int cnt = 0;
  vector<bool> used(h);
  for (int i = 0; i < h; ++i) {
    if (used[i]) continue;
    for (int j = i + 1; j < h; ++j) {
      if (used[j]) continue;
      if (s[i] == t[j]) {
        used[i] = used[j] = true;
        cnt += 2;
        break;
      }
    }
  }

  if (h - cnt >= 2) return false;

  if (h & 1) {
    for (int i = 0; i < h; ++i) {
      if (used[i]) continue;
      if (s[i] == t[i]) cnt++;
    }
  }

  return cnt == h;
}

bool solve(vector<string> &t, int q = 400000) {
  vector<string> s = t;
  vector<int> v(w);
  for (int i = 0; i < w; ++i) v[i] = i;
  while (q--) {
    //怒りの乱択
    if (check(s)) return true;
    shuffle(v.begin(), v.end(), ei);
    for (int i = 0; i < h; ++i) {
      for (int j = 0; j < w; ++j) {
        s[i][j] = t[i][v[j]];
      }
    }
  }

  return false;
}

int main() {
  cin >> h >> w;

  vector<string> s(h);
  for (auto &i : s) cin >> i;

  cout << (solve(s) ? "YES" : "NO") << endl;
}
