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 N = 1010, mod = 1e9 + 7;
vector<int> adj[N];
bool mark[N];
int a[N], power[N][N], sz = 0;
string s[N];
long long f[N], comb[N][N];
void dfs(int v) {
  sz++;
  mark[v] = true;
  for (auto u : adj[v])
    if (!mark[u]) dfs(u);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int m, n;
  cin >> m >> n;
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int b1 = 0; b1 < m; b1++) {
    for (int b2 = b1 + 1; b2 < m; b2++) {
      bool ok = true;
      for (int i = 0; i < n; i++)
        if (s[i][b1] != s[i][b2]) ok = false;
      if (ok) adj[b1].push_back(b2), adj[b2].push_back(b1);
    }
  }
  comb[0][0] = 1;
  for (int i = 1; i < N; i++) {
    comb[i][0] = comb[i][i] = 1;
    for (int j = 1; j < i; j++)
      comb[i][j] = (comb[i - 1][j] + comb[i - 1][j - 1]) % mod;
  }
  int cnt = 0;
  for (int i = 0; i < m; i++)
    if (!mark[i]) {
      sz = 0;
      dfs(i);
      a[cnt++] = sz;
    }
  f[0] = 1;
  for (int i = 1; i < N; i++) {
    for (int k = 1; k <= i; k++) {
      long long way = comb[i - 1][k - 1];
      way *= f[i - k];
      way %= mod;
      f[i] = (f[i] + way) % mod;
    }
  }
  long long ans = 1;
  for (int i = 0; i < cnt; i++) {
    ans = (1LL * ans * f[a[i]]) % mod;
  }
  cout << ans << endl;
}
