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;
int m, n;
string bs[50];
long long component[1000];
long long combo[1001][1001];
long long pseudopart[1001];
bool visited[1000];
int main() {
  cin >> m >> n;
  for (int i = 0; i < n; i++) cin >> bs[i];
  for (int i = 0; i < m; i++)
    for (int j = 0; j < n; j++) {
      component[i] *= 2LL;
      component[i] += (long long)((int)bs[j][i] - (int)'0');
    }
  combo[0][0] = 1;
  for (int i = 1; i < 1001; i++)
    for (int j = 0; j <= i; j++) {
      if (j == 0 || j == i)
        combo[i][j] = 1;
      else
        combo[i][j] = combo[i - 1][j] + combo[i - 1][j - 1];
      combo[i][j] %= 1000000007;
    }
  pseudopart[0] = 1;
  pseudopart[1] = 1;
  for (int i = 2; i <= m; i++) {
    for (int j = 0; j < i; j++) {
      pseudopart[i] += combo[i - 1][j] * pseudopart[i - 1 - j];
      pseudopart[i] %= 1000000007;
    }
  }
  long long ans = 1;
  for (int i = 0; i < m; i++) {
    if (visited[i]) continue;
    long long temp = 0;
    for (int j = 0; j < m; j++)
      if (component[j] == component[i]) {
        visited[j] = true;
        temp++;
      }
    ans *= pseudopart[temp];
    ans %= 1000000007;
  }
  cout << ans << endl;
}
