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 s[(1005)][(1005)], b[(1005)], vis[(1005)], m, n, ans;
long long bin[(1005)];
char a[(1005)];
int main() {
  cin >> m >> n, s[0][0] = ans = 1;
  for (register int i = 0; i < n; ++i) {
    scanf("%s", a);
    for (register int j = 0; j < m; ++j)
      if (a[j] == '1') bin[j] |= 1LL << i;
  }
  for (register int i = 1; i <= m; ++i)
    for (register int j = 1; j <= i; ++j) {
      s[i][j] = (1LL * s[i - 1][j] * j + s[i - 1][j - 1]) % (1000000007);
      b[i] += s[i][j];
      if (b[i] >= (1000000007)) b[i] -= (1000000007);
    }
  for (register int i = 0, sz; i < m; ++i) {
    if (vis[i]) continue;
    sz = 0;
    for (register int j = i; j < m; ++j)
      if (bin[i] == bin[j]) ++sz, vis[j] = 1;
    ans = 1LL * ans * b[sz] % (1000000007);
  }
  cout << ans;
  return 0;
}
