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 long long int M = 1000000007;
int m, n;
string T[64];
bool dis[1024];
int csz[1024];
long long int bell[1024][1024];
int main() {
  cin.sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> m >> n;
  for (int i = 0; i < n; ++i) {
    cin >> T[i];
  }
  for (int a = 0; a < m; ++a) {
    for (int b = a + 1; b < m; ++b) {
      bool diff = false;
      for (int i = 0; i < n; ++i) {
        if (T[i][a] != T[i][b]) {
          diff = true;
          break;
        }
      }
      if (!diff) {
        dis[b] = true;
        ++csz[a];
      }
    }
  }
  for (int i = 0; i < 1024; ++i) {
    if (i == 0) {
      bell[i][0] = 1;
    } else {
      bell[i][0] = bell[i - 1][i - 1];
    }
    for (int j = 1; j <= i; ++j) {
      bell[i][j] = (bell[i - 1][j - 1] + bell[i][j - 1]) % M;
    }
  }
  long long int ret = 1;
  for (int i = 0; i < m; ++i) {
    if (dis[i]) continue;
    int sz = csz[i] + 1;
    ret *= bell[sz][0];
    ret %= M;
  }
  cout << ret << '\n';
  return 0;
}
