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 mo = 1e9 + 7;
long long f[1002][1002], sum[1001], ans = 1;
int n, m, a[1001];
char s[1001];
map<long long, int> M;
int main() {
  scanf("%d%d", &m, &n);
  f[0][0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= i; j++)
      sum[i] =
          (sum[i] + (f[i][j] = (f[i - 1][j] * j + f[i - 1][j - 1]) % mo)) % mo;
  for (int i = 0; i < n && ~scanf("%s", s); i++)
    for (int j = 0; j < m; j++) a[j] += (s[j] - 48) << i;
  for (int i = 0; i < m; i++) M[a[i]]++;
  for (int i = 1; i < m; i++)
    if (M[a[i]]) ans = ans * sum[M[a[i]]] % mo, M[a[i]] = 0;
  printf("%lld\n", ans);
  return 0;
}
