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;
int n, m, i, j;
char s[52][1024];
int ord[1024];
long long comb[1024][1024];
long long bell[1024];
long long ans;
int main() {
  scanf("%d%d\n", &m, &n);
  for (i = 1; i <= n; i++) scanf("%s\n", s[i] + 1);
  for (i = 1; i <= m; i++) ord[i] = i;
  auto cmp = [](int a, int b) -> int const {
    int i;
    for (i = 1; i <= n; i++)
      if (s[i][a] != s[i][b]) break;
    if (i > n) return 0;
    return (s[i][a] < s[i][b] ? 1 : 0);
  };
  sort(ord + 1, ord + m + 1, cmp);
  bell[0] = 1;
  for (i = 0; i <= m; i++) {
    comb[i][0] = 1;
    for (j = 1; j <= i; j++) {
      comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];
      if (comb[i][j] >= 1000000007) comb[i][j] -= 1000000007;
      bell[i] = (bell[i] + comb[i - 1][j - 1] * bell[i - j]) % 1000000007;
    }
  }
  ans = 1LL;
  for (i = 1; i <= m; i = j) {
    for (j = i; cmp(ord[i], ord[j]) == 0 && cmp(ord[j], ord[i]) == 0; j++)
      ;
    ans *= bell[j - i];
    ans %= 1000000007;
  }
  cout << ans;
  return 0;
}
