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 q = 1000000007;
int m, n, x[1010], f[1010], c[1010][1010], p;
char s[60][1010];
int main() {
  int i, j, k;
  scanf("%d%d", &m, &n);
  for (i = 1; i <= n; i++) scanf("%s", &s[i]);
  for (i = 0; i < m; i++) {
    for (j = 0; j < i; j++) {
      for (k = 1; k <= n; k++)
        if (s[k][i] != s[k][j]) break;
      if (k > n) break;
    }
    x[j]++;
  }
  for (i = 0; i <= m; i++)
    for (c[i][0] = 1, j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % q;
  f[0] = f[1] = 1;
  for (i = 2; i <= m; i++)
    for (j = 1; j <= i; j++)
      f[i] = (f[i] + (long long)c[i - 1][j - 1] * f[i - j]) % q;
  p = 1;
  for (i = 0; i < m; i++) p = (long long)p * f[x[i]] % q;
  cout << p;
  return 0;
}
