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 inv = 1000000000;
const int minv = -inv;
const int max_m = 1000 + 5;
const int max_n = 50 + 5;
const long long modref = ((long long)(1e9 + 7));
long long s[max_m + 1][max_m + 1] = {0};
long long q[max_m + 1];
int m, n;
char e[max_n][max_m];
map<string, int> M;
map<string, int>::iterator it;
int main() {
  scanf("%d", &m);
  scanf("%d", &n);
  q[0] = 1ll;
  for (int i = 1; i <= m; ++i) {
    s[i][1] = 1ll;
    q[i] = 1ll;
    for (int j = 2; j <= i; ++j) {
      s[i][j] = (s[i - 1][j] * ((long long)(j)) + s[i - 1][j - 1]) % modref;
      q[i] = (q[i] + s[i][j]) % modref;
    }
  }
  M.clear();
  for (int i = 0; i < n; ++i) scanf("%s", e[i]);
  for (int j = 0; j < m; ++j) {
    string curs = "";
    for (int i = 0; i < n; ++i) curs += e[i][j];
    ++M[curs];
  }
  long long res = 1ll;
  for (it = M.begin(); it != M.end(); ++it)
    res = (res * q[it->second]) % modref;
  cout << res << "\n";
  return 0;
}
