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;
long long ans, c[1002][1002], f[1002];
int nu[1002], m, n, a[52][1002], b[1002][1002];
bool comp(int _, int __) {
  for (int i = 1; i <= m; ++i)
    if (b[_][i] != b[__][i]) return b[_][i] < b[__][i];
  return 0;
}
bool equ(int _, int __) {
  for (int i = 1; i <= m; ++i)
    if (b[_][i] != b[__][i]) return 0;
  return 1;
}
int main() {
  scanf("%d%d\n", &m, &n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) scanf("%1d", &a[i][j]);
  }
  for (int j = 1; j <= m; ++j) {
    for (int i = 1; i <= m; ++i) b[j][i] = 1;
    for (int i = 1; i <= n; ++i) {
      if (a[i][j])
        for (int k = 1; k <= m; ++k) b[j][k] &= a[i][k];
      else
        for (int k = 1; k <= m; ++k) b[j][k] &= (1 - a[i][k]);
    }
  }
  for (int i = 1; i <= m; ++i) nu[i] = i;
  sort(nu + 1, nu + m + 1, comp);
  c[0][0] = 1;
  for (int i = 1; i <= m; ++i)
    for (int j = 0; j <= i; ++j)
      c[i][j] = j ? (c[i - 1][j - 1] + c[i - 1][j]) % ((int)1e9 + 7) : 1;
  f[0] = f[1] = 1;
  for (int i = 2; i <= m; ++i)
    for (int j = 0; j < i; ++j)
      f[i] = (f[i] + c[i - 1][j] * f[j] % ((int)1e9 + 7)) % ((int)1e9 + 7);
  ans = 1;
  for (int i = 1; i <= m;) {
    int t = i;
    while (i < m && equ(nu[i], nu[i + 1])) i++;
    ans = (ans * f[i - t + 1]) % ((int)1e9 + 7);
    i++;
  }
  printf("%I64d", ans);
  return 0;
}
