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 = 1000000007;
int n, m;
char s[1005];
long long a[1005];
int f[1005][1005];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (int)(1); i <= (int)(m); i++) {
    scanf("%s", s + 1);
    for (int j = (int)(1); j <= (int)(n); j++)
      if (s[j] == '1') a[j] |= 1ll << i;
  }
  f[1][1] = 1;
  for (int i = (int)(2); i <= (int)(1000); i++)
    for (int j = (int)(1); j <= (int)(i); j++)
      f[i][j] = (1ll * j * f[i - 1][j] + f[i - 1][j - 1]) % mo;
  sort(a + 1, a + n + 1);
  int la = 1, ans = 1;
  for (int i = (int)(2); i <= (int)(n + 1); i++)
    if (a[i] == a[i - 1] && i != n + 1)
      ++la;
    else {
      int s = 0;
      for (int j = (int)(1); j <= (int)(la); j++) s = (s + f[la][j]) % mo;
      ans = 1ll * ans * s % mo;
      la = 1;
    }
  cout << ans << endl;
}
