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;
string t[100];
int n, m;
int64_t C[1001][1001];
int64_t Bell[1001];
const int64_t M = 1000000007;
void prepare_bell() {
  for (int i = 0; i <= 1000; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % M;
  }
  Bell[0] = Bell[1] = 1;
  for (int i = 2; i <= 1000; i++) {
    int64_t sum = 0;
    for (int j = 0; j <= i - 1; j++) sum = (sum + C[i - 1][j] * Bell[j]) % M;
    Bell[i] = sum;
  }
}
bool equiv(int x, int y) {
  for (int i = 0; i < n; i++) {
    if (t[i][x] != t[i][y]) return false;
  }
  return true;
}
int main() {
  prepare_bell();
  cin >> m >> n;
  for (int i = 0; i < n; i++) cin >> t[i];
  vector<bool> used(m, false);
  int64_t ans = 1;
  for (int i = 0; i < m; i++) {
    if (used[i]) continue;
    used[i] = true;
    vector<int> f;
    f.push_back(i);
    for (int j = i + 1; j < m; j++) {
      if (equiv(i, j)) {
        used[j] = true;
        f.push_back(j);
      }
    }
    ans = (ans * Bell[f.size()]) % M;
  }
  cout << ans << "\n";
  return 0;
}
