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 <algorithm>
#include <cstdio>
#include <cstring>

const int N = 15;

int n, m;
char s[N][N];

int p[N][2], cnt;

bool vis[N];

bool check() {
  int t = n / 2, ss = 0;
  memset(vis, 0, sizeof vis);
  for (int i = 0; i < m; ++i) if (!vis[i]) {
    bool ok;
    for (int j = i + 1; j < m; ++j) if (!vis[j]) {
      ok = true;
      for (int k = 0; k < t; ++k)
        if (s[p[k][0]][i] != s[p[k][1]][j] || s[p[k][0]][j] != s[p[k][1]][i]) {
          ok = false;
          break;
        }
      if (ok) {
        vis[j] = true;
        break;
      }
    }
    if (!ok) {
      for (int k = 0; k < t; ++k)
        if (s[p[k][0]][i] != s[p[k][1]][i])
          return false;
      ++ss;
    }
  }
  return ss <= m % 2;
}

bool dfs(int S, bool t) {
  if (S == 0) return check();
  if (t && dfs(S & (S - 1), false)) return true;
  if (S != (S & -S)) {
    int x = 0;
    while ((~S >> x) & 1) ++x;
    for (int y = x + 1; y < n; ++y) if ((S >> y) & 1) {
      p[cnt][0] = x; p[cnt][1] = y;
      ++cnt;
      if (dfs(S & ~(1 << x) & ~(1 << y), t)) return true;
      --cnt;
    }
  }
  return false;
}

int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i)
    scanf("%s", s[i]);
  puts(dfs((1 << n) - 1, n % 2 == 1) ? "YES" : "NO");
  return 0;
}
