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;
long long int MOD = 1000000007;
template <typename T>
vector<T> get_vector(long long int size) {
  vector<T> out;
  for (long long int sahdfjhfsk = 0; sahdfjhfsk < size; ++sahdfjhfsk) {
    T temp;
    cin >> temp;
    out.push_back(temp);
  }
  return out;
}
vector<vector<long long int>> bell_trig = {{1}};
long long int bell(long long int n) {
  if (bell_trig.size() > n) return bell_trig[n][0];
  bell(n - 1);
  bell_trig.push_back({bell_trig.back().back()});
  for (long long int i = 1; i <= n; ++i) {
    bell_trig.back().push_back((bell_trig[n][i - 1] + bell_trig[n - 1][i - 1]) %
                               MOD);
  }
  return bell(n);
}
int main() {
  long long int m, n;
  cin >> m >> n;
  vector<vector<bool>> ns;
  for (long long int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    ns.emplace_back();
    for (char c : s) ns.back().push_back(c == '1');
  }
  vector<vector<long long int>> primes{{}};
  for (long long int i = 0; i < m; ++i) primes.back().push_back(i);
  for (vector<bool> &num : ns) {
    vector<vector<long long int>> nprimes;
    for (vector<long long int> &subset : primes) {
      nprimes.emplace_back();
      for (long long int s : subset)
        if (num[s]) {
          nprimes.back().push_back(s);
        }
      if (nprimes.back().empty()) nprimes.pop_back();
      nprimes.emplace_back();
      for (long long int s : subset)
        if (!num[s]) {
          nprimes.back().push_back(s);
        }
      if (nprimes.back().empty()) nprimes.pop_back();
    }
    swap(primes, nprimes);
  }
  long long int total = 1;
  for (auto &p : primes) {
    total *= bell(p.size());
    total = total % MOD;
  }
  cout << total << endl;
}
