You are given an integer m.

Let M = 2m - 1.

You are also given a set of n integers denoted as the set T. The integers will be provided in base 2 as n binary strings of length m.

A set of integers S is called "good" if the following hold. 

  1. If <image>, then <image>. 
  2. If <image>, then <image>
  3. <image>
  4. All elements of S are less than or equal to M. 



Here, <image> and <image> refer to the bitwise XOR and bitwise AND operators, respectively.

Count the number of good sets S, modulo 109 + 7.

Input

The first line will contain two integers m and n (1 ≤ m ≤ 1 000, 1 ≤ n ≤ min(2m, 50)).

The next n lines will contain the elements of T. Each line will contain exactly m zeros and ones. Elements of T will be distinct.

Output

Print a single integer, the number of good sets modulo 109 + 7. 

Examples

Input

5 3
11010
00101
11000


Output

4


Input

30 2
010101010101010010101010101010
110110110110110011011011011011


Output

860616440

Note

An example of a valid set S is {00000, 00101, 00010, 00111, 11000, 11010, 11101, 11111}.
#include <bits/stdc++.h>
using namespace std;
const int p = 1000000007;
int m, n;
int mp[2005][2005], x[2005][2005], f[2005][2005], fa[2005], size[2005];
int find(int x) {
  if (x == fa[x]) return x;
  return fa[x] = find(fa[x]);
}
char s[2005];
int main() {
  cin >> m >> n;
  f[0][0] = 1;
  for (int i = 1; i <= 2000; i++) {
    fa[i] = i;
    size[i] = 1;
    for (int j = 1; j <= i; j++) {
      f[i][j] = (f[i - 1][j - 1] + 1LL * j * f[i - 1][j] % p) % p;
    }
  }
  for (int i = 1; i <= n; i++) {
    scanf("%s", s + 1);
    for (int j = 1; j <= m; j++) {
      x[i][j] = s[j] - '0';
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      for (int k = 1; k < j; k++) {
        if (x[i][j] != x[i][k]) mp[k][j] = 1;
      }
    }
  }
  int ans = 1;
  for (int i = 1; i < m; i++) {
    for (int j = i + 1; j <= m; j++) {
      if (!mp[i][j]) {
        int aa = find(i), bb = find(j);
        if (aa != bb) {
          size[bb] += size[aa];
          fa[aa] = bb;
        }
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    if (find(i) == i) {
      int tmp = 0;
      for (int j = 1; j <= size[i]; j++) (tmp += f[size[i]][j]) %= p;
      ans = 1LL * ans * tmp % p;
    }
  }
  cout << ans << endl;
  return 0;
}
