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 maxk = 1005;
const long long M = 1e9 + 7;
long long s[maxk][maxk];
long long vt[maxk][maxk];
string pat[maxk];
long long b[maxk];
int main() {
  long long m, n;
  cin >> m >> n;
  for (long long i = 0; i <= m; i++) s[i][i] = 1, s[i][1] = 1;
  for (long long i = 2; i <= m; i++) {
    for (long long j = 2; j < i; j++)
      s[i][j] = (j * s[i - 1][j] % M + s[i - 1][j - 1]) % M;
  }
  for (long long i = 0; i < n; i++) cin >> pat[i];
  map<string, long long> ss;
  map<string, long long>::iterator it;
  long long val = m;
  for (int i = 0; i < m; i++) {
    string bat = "";
    for (int j = 0; j < m; j++) bat += "1";
    for (long long j = 0; j < n; j++)
      for (long long k = 0; k < m; k++) {
        char gg = pat[j][i] == '1' ? pat[j][k] : (pat[j][k] == '0' ? '1' : '0');
        bat[k] = bat[k] == '0' ? '0' : gg;
      }
    if (ss.find(bat) != ss.end()) continue;
    long long sz = 0;
    for (long long j = 0; j < m; j++)
      if (bat[j] == '1') sz++, val--;
    ss[bat] = sz;
  }
  for (long long i = 0; i <= m; i++) {
    long long aux = 0;
    for (long long j = 0; j <= i; j++) aux = (aux + s[i][j]) % M;
    b[i] = aux;
  }
  long long ans = b[val];
  for (it = ss.begin(); it != ss.end(); it++) ans = (ans * b[it->second]) % M;
  cout << ans;
}
