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 N = 1111;
const int mod = 1e9 + 7;
int n, c[N][N], f[N];
char ch[N];
int m;
long long a[N];
int main() {
  int i, j, ans = 1;
  cin >> n >> m;
  for (i = 0; i <= n; i = i + 1) {
    c[i][0] = 1, c[i][i] = 1;
    for (j = 1; j < i; j = j + 1)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
  }
  f[0] = 1;
  for (i = 1; i <= n; i = i + 1) {
    f[i] = 0;
    for (j = 1; j <= i; j = j + 1)
      f[i] = (f[i] + (long long)c[i - 1][j - 1] * f[i - j] % mod) % mod;
  }
  for (i = 1; i <= m; i = i + 1) {
    cin >> ch;
    for (j = 1; j <= n; j = j + 1)
      a[j] = (long long)a[j] * 2 + (long long)(ch[j - 1] - 48);
  }
  sort(a + 1, a + n + 1);
  j = 1;
  for (i = 2; i <= n + 1; i = i + 1) {
    if (i <= n && a[i] == a[i - 1])
      j++;
    else
      ans = (long long)ans * f[j] % mod, j = 1;
  }
  cout << ans << endl;
  return 0;
}
