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 P = 1e9 + 7;
char rs[(1000 + 5)];
int f[(1000 + 5)][(1000 + 5)], s[(1000 + 5)], zt[(1000 + 5)], g[(1000 + 5)];
int main() {
  int m, n;
  scanf("%d%d", &m, &n);
  for (int i = (1); i <= (int)(n); ++i) {
    scanf("%s", rs + 1);
    int len = strlen(rs + 1);
    for (int j = (1); j <= (int)(len); ++j)
      if (rs[j] == '1') zt[j] |= 1ll << i;
  }
  f[0][0] = s[0] = 1;
  for (int i = (1); i <= (int)(m); ++i)
    for (int j = (1); j <= (int)(i); ++j) {
      f[i][j] = ((1ll * f[i - 1][j] * j % P) + f[i - 1][j - 1]) % P;
      s[i] = (s[i] + f[i][j]) % P;
    }
  int ans = 1;
  for (int i = (1); i <= (int)(m); ++i)
    if (!g[i]) {
      int siz = 0;
      for (int j = (i); j <= (int)(m); ++j)
        if (zt[i] == zt[j]) ++siz, g[j] = i;
      ans = 1ll * ans * s[siz] % P;
    }
  printf("%d\n", ans);
  return 0;
}
