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>
const int mo = 1e9 + 7;
using namespace std;
char s[55][1050];
int bl[1050], tot, n, m, g[1050][1050], f[1050], mark[1050], ans = 1;
int same(int x, int y) {
  for (int i = 1; i <= n; ++i)
    if (s[i][x] != s[i][y]) return 0;
  return 1;
}
int main() {
  cin >> m >> n;
  for (int i = 1; i <= n; ++i) scanf("%s", s[i]);
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < i; ++j)
      if (same(j, i)) {
        bl[i] = bl[j];
        break;
      }
    if (!bl[i]) bl[i] = ++tot;
  }
  g[0][0] = f[0] = 1;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= i; ++j)
      g[i][j] = (1ll * g[i - 1][j] * j + g[i - 1][j - 1]) % mo,
      (f[i] += g[i][j]) %= mo;
  for (int i = 0; i < m; ++i) ++mark[bl[i]];
  for (int i = 0; i < m; ++i) ans = 1ll * ans * f[mark[i]] % mo;
  cout << ans;
  return 0;
}
