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 = 1005;
const int MOD = 1E9 + 7;
int n, m, sum;
long long f[N], C[N][N], ans, g[N];
char s[N];
int main() {
  cin >> m >> n;
  for (int i = 0; i <= m; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
  }
  f[0] = ans = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < i; j++)
      f[i] = (f[i] + C[i - 1][j] * f[i - 1 - j]) % MOD;
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    for (int j = 0; j < m; j++)
      if (s[j] == '1') g[j] += (long long)1 << i;
  }
  sort(g, g + m);
  g[m] = -1;
  for (int i = 1; i <= m; i++)
    if (g[i] == g[i - 1])
      sum++;
    else {
      ans = ans * f[sum + 1] % MOD;
      sum = 0;
    }
  cout << ans;
}
