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 md = 1e9 + 7;
const int maxn = 1006;
string s[maxn];
long long B[maxn];
long long C[maxn][maxn];
map<string, int> M;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int m, n;
  cin >> m >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  for (int i = 0; i < maxn; i++) {
    for (int j = 0; j < maxn; j++) {
      if (j == 0) {
        C[i][j] = 1;
        continue;
      }
      if (i == 0) {
        C[i][j] = 0;
        continue;
      }
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % md;
    }
  }
  B[0] = B[1] = 1;
  for (int i = 2; i < maxn; i++) {
    B[i] = 0;
    for (int j = 0; j < i; j++) {
      B[i] = (B[i] + (B[j] * C[i - 1][j]) % md) % md;
    }
  }
  for (int i = 0; i < m; i++) {
    string t = "";
    for (int j = 1; j <= n; j++) t += s[j][i];
    if (M.count(t) == 0) M[t] = 0;
    M[t]++;
  }
  long long ans = 1;
  for (auto elem : M) {
    ans = (ans * B[elem.second]) % md;
  }
  cout << ans << endl;
}
