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 main() {
  int m, n;
  cin >> m >> n;
  vector<string> nums(n);
  for (int i = 0; i < (int)(n); ++i) cin >> nums[i];
  vector<long long> c(m, 1);
  for (int i = 0; i < (int)(m); ++i) {
    for (int j = 0; j < (int)(i); ++j) {
      int match = 1;
      for (int k = 0; k < (int)(n); ++k) {
        if (nums[k][i] != nums[k][j]) {
          match = 0;
          break;
        }
      }
      if (match) {
        c[i] = 0;
        c[j]++;
        break;
      }
    }
  }
  vector<vector<long long> > choose(m + 1, vector<long long>(m + 1, 0));
  for (int i = 0; i < (int)(m + 1); ++i) {
    choose[i][0] = 1;
    choose[i][i] = 1;
    for (int j = 0; j < (int)(m + 1); ++j) {
      if (i && j)
        choose[i][j] = (choose[i - 1][j] + choose[i - 1][j - 1]) % 1000000007;
    }
  }
  vector<long long> dp(m + 1);
  dp[0] = 1;
  for (int i = 0; i < (int)(m + 1); ++i) {
    if (i) {
      for (int j = 0; j < (int)(i); ++j) {
        dp[i] += (choose[i - 1][j] * dp[i - 1 - j]) % 1000000007;
        dp[i] %= 1000000007;
      }
    }
  }
  long long ans = 1;
  for (int i = 0; i < (int)(m); ++i) {
    if (c[i]) ans = (ans * dp[c[i]]) % 1000000007;
  }
  cout << ans << endl;
  return 0;
}
