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;
int m, n;
int R[1005];
map<int, int> cnt;
long long bell[1005][1005];
void geneBell() {
  bell[0][0] = 1;
  for (int i = 1; i < m + 2; i++) {
    bell[i][0] = bell[i - 1][i - 1];
    for (int j = 1; j <= i; j++) {
      bell[i][j] = (bell[i][j - 1] + bell[i - 1][j - 1]) % 1000000007;
    }
  }
}
int main() {
  cin >> m >> n;
  geneBell();
  string tmp;
  for (int i = 0; i < n; i++) {
    cin >> tmp;
    for (int j = 0; j < m; j++) {
      R[j] = R[j] * 10 + tmp[j] - '0';
    }
  }
  for (int i = 0; i < m; i++) {
    cnt[R[i]]++;
  }
  map<int, int>::iterator it;
  long long ans = 1;
  for (it = cnt.begin(); it != cnt.end(); ++it) {
    if ((it->second) > 0) {
      ans = (ans * bell[it->second][0]) % 1000000007;
    }
  }
  cout << ans;
  return 0;
}
