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 MM = 1e9 + 7;
const double eps = 1e-8;
const int MAXN = 2e6 + 10;
int n, m;
void prework() {}
void read() {}
long long f[1024];
char s[1024];
long long a[1024][1024];
void solve(int casi) {
  scanf("%d%d", &m, &n);
  map<long long, int> mp;
  for (int i = 1; i <= n; i++) {
    scanf("%s", s);
    for (int j = 1; j <= m; j++) {
      f[j] = f[j] * 2 + (s[j - 1] - '0');
    }
  }
  for (int j = 1; j <= m; j++) mp[f[j]]++;
  a[1][1] = 1;
  for (int i = 2; i <= m + 1; i++) {
    a[i][1] = a[i - 1][i - 1];
    for (int j = 2; j <= i; j++) a[i][j] = (a[i][j - 1] + a[i - 1][j - 1]) % MM;
  }
  long long ans = 1;
  for (auto &x : mp) ans = (ans * a[x.second + 1][1]) % MM;
  cout << ans << endl;
}
void printans() {}
int main() {
  prework();
  int T = 1;
  for (int i = 1; i <= T; i++) {
    read();
    solve(i);
    printans();
  }
  return 0;
}
