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 MOD = 1000000007;
void yes() {
  cout << "Yes"
       << "\n";
}
void no() {
  cout << "No"
       << "\n";
}
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
struct FAIL {
  FAIL() {
    cout << "CHANGE!!!"
         << "\n";
  }
};
vector<vector<long long>> calc_stirling_2(int n) {
  vector<vector<long long>> res(n + 1, vector<long long>(n + 1));
  res[0][0] = 1;
  for (int i = (1); i < (n + 1); ++i) {
    for (int j = (1); j < (n + 1); ++j) {
      res[i][j] = res[i - 1][j - 1] + j * res[i - 1][j];
      res[i][j] %= MOD;
    }
  }
  return res;
}
vector<long long> calc_bell(int n) {
  auto st = calc_stirling_2(n);
  vector<long long> res(n + 1);
  for (int i = (0); i < (n + 1); ++i) {
    for (int j = (0); j < (n + 1); ++j) {
      res[i] += st[i][j];
      res[i] %= MOD;
    }
  }
  return res;
}
const string inv(string s) {
  for (int i = (0); i < (s.size()); ++i) {
    s[i] = s[i] == '1' ? '0' : '1';
  }
  return s;
}
int main() {
  int n, m;
  cin >> m >> n;
  vector<bitset<1000>> v;
  for (int i = (0); i < (n); ++i) {
    string s;
    cin >> s;
    v.push_back(bitset<1000>{s});
    v.push_back(bitset<1000>{inv(s)});
  }
  n *= 2;
  vector<bitset<1000>> minc(m, bitset<1000>{string(1000, '1')});
  for (int i = (0); i < (m); ++i) {
    for (int j = (0); j < (n); ++j) {
      if (v[j].test(i)) {
        minc[i] &= v[j];
      }
    }
  }
  unordered_set<bitset<1000>> minc_hash = {minc.begin(), minc.end()};
  auto bell = calc_bell(m);
  long long res = 1;
  for (const auto &mask : minc_hash) {
    res *= bell[mask.count()];
    res %= MOD;
  }
  cout << res;
}
