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;
string s;
int cnt[1000][1000];
int m;
int n;
int size[1000];
long long c[1001][1001];
long long way[1001];
long long p2[1001];
int main() {
  ios::sync_with_stdio(false);
  cin >> m >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s;
    for (int j = 0; j < m; j++) {
      for (int k = 0; k < m; k++) {
        cnt[j][k] += (s[j] == s[k]);
      }
    }
  }
  c[0][0] = 1;
  way[0] = 1;
  p2[0] = 1;
  for (int i = 1; i <= m; i++) {
    p2[i] = p2[i - 1] * 2 % 1000000007;
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
      c[i][j] %= 1000000007;
    }
    way[i] = 0;
    for (int j = 0; j < i; j++) {
      way[i] = (way[i] + way[i - j - 1] * c[i - 1][j]);
      way[i] %= 1000000007;
    }
  }
  int sets = 0;
  for (int i = 0; i < m; i++) {
    bool eq = false;
    for (int j = 0; j < i; j++) {
      if (cnt[i][j] == n) {
        size[j]++;
        break;
      }
    }
    if (!eq) size[i] = 1;
  }
  long long ans = 1;
  for (int i = 0; i < m; i++) {
    if (size[i] == 0) continue;
    ans *= way[size[i]];
    ans %= 1000000007;
  }
  cout << ans << endl;
}
