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 long long mod = 1e9 + 7;
long long mask[1020], f[1020][1020], sum[1020];
char ch[1020];
int n, m, vis[1020];
void init() {
  f[0][0] = 1;
  for (register int i = 1; i <= m; i++) {
    for (register int j = 1; j <= i; j++) {
      f[i][j] = (f[i - 1][j - 1] + f[i - 1][j] * j) % mod;
      sum[i] = (sum[i] + f[i][j]) % mod;
    }
  }
}
int main() {
  scanf("%d %d", &m, &n);
  init();
  for (register int i = 1; i <= n; i++) {
    scanf("%s", ch + 1);
    for (register int j = 1; j <= m; j++) {
      mask[j] |= (1ll << (i - 1)) * (ch[j] - '0');
    }
  }
  long long ans = 1;
  for (register int i = 1; i <= m; i++) {
    if (!vis[i]) {
      vis[i] = 1;
      int sz = 1;
      for (register int j = 1; j <= m; j++) {
        if (vis[j]) continue;
        if (mask[i] == mask[j]) vis[j] = 1, sz++;
      }
      ans = ans * sum[sz] % mod;
    }
  }
  cout << ans << endl;
}
