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, P = 1e9 + 7;
int ans = 1, B[N], C[N][N], m, n;
bitset<N> vis, f, b[60], t;
char s[N];
int main() {
  scanf("%d%d", &m, &n);
  B[0] = 1;
  for (int i = (1); i <= (m); ++i) f[i] = 1;
  for (int i = (0); i <= (m); ++i) {
    C[i][0] = 1;
    for (int j = (1); j <= (i); ++j)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
  }
  for (int i = (1); i <= (m); ++i)
    for (int j = (0); j <= (i - 1); ++j)
      B[i] = (B[i] + 1ll * B[j] * C[i - 1][j]) % P;
  for (int i = (1); i <= (n); ++i) {
    scanf("%s", s + 1);
    for (int j = (1); j <= (m); ++j) b[i][j] = s[j] - 48;
  }
  for (int i = (1); i <= (m); ++i)
    if (!vis[i]) {
      t = f;
      for (int j = (1); j <= (n); ++j)
        if (b[j][i])
          t &= b[j];
        else
          t &= f ^ b[j];
      ans = 1ll * ans * B[t.count()] % P;
      vis |= t;
    }
  printf("%d\n", ans);
  return 0;
}
