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 MAXN = 1010, MOD = 1e9 + 7;
int m, n;
string a[51];
bool link[MAXN][MAXN], cho[MAXN];
int p2[MAXN], f[MAXN][MAXN], g[MAXN];
int main() {
  ios::sync_with_stdio(0);
  cin >> m >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  memset(link, 1, sizeof link);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < m; ++j)
      for (int t = j + 1; t < m; ++t)
        if (a[i][j] != a[i][t]) link[j + 1][t + 1] = link[t + 1][j + 1] = 0;
  }
  f[0][0] = 1;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= i; ++j) {
      f[i][j] = (f[i - 1][j - 1] + 1ll * f[i - 1][j] * j) % MOD;
      g[i] = (g[i] + f[i][j]) % MOD;
    }
  long long ans = 1;
  for (int i = 1; i <= m; ++i)
    if (!cho[i]) {
      int now = 0;
      for (int j = i; j <= m; ++j)
        if (link[i][j]) {
          cho[j] = 1;
          now++;
        }
      ans = ans * g[now] % MOD;
    }
  cout << ans << endl;
  return 0;
}
