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 n, m;
string a[1007];
long long mask[1007];
long long dp[1007][1007];
long long hh[1007];
void input() {
  cin >> m >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    for (int j = 0; j < m; ++j) {
      if (a[i][j] == '1') {
        mask[j] += (1LL << i);
      }
    }
  }
}
void solve() {
  dp[0][0] = 1;
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j <= i; ++j) {
      dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j]) % 1000000007;
      dp[i + 1][j] = (dp[i + 1][j] + (dp[i][j] * j) % 1000000007) % 1000000007;
    }
  }
  for (int i = 1; i <= m; ++i) {
    for (int j = 1; j <= i; ++j) {
      hh[i] = (hh[i] + dp[i][j]) % 1000000007;
    }
  }
  sort(mask, mask + m);
  long long ans = 1;
  int len = 1;
  for (int i = 1; i < m; ++i) {
    if (mask[i] == mask[i - 1]) {
      ++len;
    } else {
      ans = (ans * hh[len]) % 1000000007;
      len = 1;
    }
  }
  ans = (ans * hh[len]) % 1000000007;
  cout << ans << "\n";
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}
