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;
long long comb[2000][2000], part[2000];
int cnt[2000], f[2000];
char c[50][2000];
long long modu = 1000000007;
int main() {
  int m, n, i, j, k;
  scanf("%d%d", &m, &n);
  for (i = 0; i < n; i++) {
    scanf("%s", c[i]);
  }
  for (i = 0; i <= m; i++) {
    for (j = 0; j <= i; j++) {
      if (i >= 1 && j >= 1)
        comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % modu;
      else
        comb[i][j] = 1;
    }
  }
  part[0] = 1;
  for (i = 1; i <= m; i++) {
    part[i] = 0;
    for (j = 0; j < i; j++) {
      part[i] += comb[i - 1][j] * part[i - 1 - j] % modu;
      part[i] %= modu;
    }
  }
  for (i = 0; i < m; i++) f[i] = i;
  bool same;
  for (i = 0; i < m; i++) {
    for (j = i + 1; j < m; j++) {
      same = true;
      for (k = 0; k < n; k++) {
        if (c[k][i] != c[k][j]) {
          same = false;
          break;
        }
      }
      if (same) {
        f[i] = f[j] = min(f[i], f[j]);
      }
    }
  }
  for (i = 0; i < m; i++) cnt[i] = 0;
  for (i = 0; i < m; i++) cnt[f[i]]++;
  long long ans = 1;
  for (i = 0; i < m; i++) {
    if (cnt[i] == 0) continue;
    ans = ans * part[cnt[i]] % modu;
  }
  printf("%lld\n", ans);
  return 0;
}
