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 mod = 1e9 + 7;
int ncr[1002][1002], bell[1002], sz[1002], par[1002];
int root(int x) { return (par[x] == x) ? x : par[x] = root(par[x]); }
int main() {
  int n, m, ans = 1;
  char s[55][1002];
  scanf("%d%d", &m, &n);
  for (int i = 0; i <= m; i++) {
    ncr[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      ncr[i][j] = (ncr[i - 1][j - 1] + ncr[i - 1][j]) % mod;
    }
  }
  bell[0] = 1;
  for (int i = 1; i <= m; i++) {
    par[i] = i;
    for (int j = 0; j < i; j++) {
      bell[i] += (1LL * ncr[i - 1][j] * bell[j]) % mod;
      bell[i] %= mod;
    }
  }
  for (int i = 1; i <= n; i++) {
    cin >> s[i] + 1;
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j < i; j++) {
      int cnt = 0;
      for (int k = 1; k <= n; k++) {
        if (s[k][i] == s[k][j]) {
          cnt++;
        }
      }
      if (cnt != n) continue;
      par[root(i)] = par[root(j)];
    }
  }
  for (int i = 1; i <= m; i++) sz[root(i)]++;
  for (int i = 1; i <= m; i++) {
    ans = (1LL * ans * bell[sz[i]]) % mod;
  }
  printf("%d\n", ans);
  return 0;
}
