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 dp[1001][1001];
const int mod = 1e9 + 7;
int f(const int n, const int k = 0) {
  if (dp[n][k] != -1) return dp[n][k];
  if (n == 0) return 1;
  return dp[n][k] = ((long long)f(n - 1, k + 1) +
                     (long long)k * (long long)f(n - 1, k)) %
                    mod;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(dp, -1, sizeof dp);
  int m, n;
  cin >> m >> n;
  vector<set<int>> sets(m);
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    for (int j = 0; j < m; ++j)
      if (s[j] == '1') sets[j].insert(i);
  }
  map<set<int>, int> occ;
  for (auto &&x : sets) ++occ[x];
  int ans = 1;
  for (auto &&x : occ) ans = ((long long)ans * (long long)f(x.second)) % mod;
  cout << ans;
  return 0;
}
