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 P = 1e9 + 7;
int n, m;
int f[1010];
vector<int> ff[1010];
string s[1010];
long long c[1010][1010];
long long B[1010];
bool same(int x, int y) {
  for (int i = 0; i < m; i++)
    if (s[i][x] != s[i][y]) return false;
  return true;
}
int main() {
  c[0][0] = 1;
  for (int i = 1; i < 1005; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      c[i][j] %= P;
    }
  }
  B[0] = 1;
  for (int i = 1; i < 1005; i++) {
    for (int j = 0; j < i; j++) {
      B[i] += B[j] * c[i - 1][j] % P;
      B[i] %= P;
    }
  }
  cin >> n >> m;
  for (int i = 0; i < m; i++) cin >> s[i];
  for (int i = 0; i < n; i++) f[i] = -1;
  for (int i = 0; i < n; i++) {
    if (f[i] == -1) {
      f[i] = i;
      ff[i].push_back(i);
      for (int j = i + 1; j < n; j++) {
        if (same(i, j)) {
          f[j] = i;
          ff[i].push_back(j);
        }
      }
    }
  }
  long long s = 1;
  for (int i = 0; i < n; i++) {
    int k = (int)ff[i].size();
    s *= B[k];
    s %= P;
  }
  cout << s << endl;
  return 0;
}
