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 = (int)1e9 + 7;
int m, n;
string a[55];
int mult(int a, int b) { return a * (long long)b % mod; }
int sum(int a, int b) {
  int ans = a + b;
  if (ans >= mod) ans -= mod;
  return ans;
}
int dp[1005][1005];
int ss[1005];
int main() {
  srand(123);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> m >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<int> dd;
  for (int i = 0; i < m; i++) dd.push_back(i);
  vector<vector<int> > t;
  t.push_back(dd);
  for (int i = 0; i < n; i++) {
    vector<vector<int> > gg;
    for (int j = 0; j < t.size(); j++) {
      vector<int> f[2];
      for (int first : t[j]) f[a[i][first] - '0'].push_back(first);
      if (f[0].size()) gg.push_back(f[0]);
      if (f[1].size()) gg.push_back(f[1]);
    }
    t = gg;
  }
  dp[1][1] = 1;
  for (int i = 1; i < 1005; i++) {
    dp[i][0] = 0;
    for (int j = 1 + (i == 1); j <= i; j++) {
      dp[i][j] = sum(dp[i - 1][j - 1], mult(j, dp[i - 1][j]));
    }
  }
  for (int i = 1; i < 1005; i++) {
    for (int j = 1; j <= i; j++) ss[i] = sum(ss[i], dp[i][j]);
  }
  int ans = 1;
  for (auto first : t) ans = mult(ans, ss[first.size()]);
  cout << ans;
}
