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 M = 1000000007;
int m, n;
char s[65][1015];
long long a[1015], b[1015], c[1015][1015];
int main() {
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; i++) scanf("%s", s[i]);
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) a[i] = a[i] * 2 + s[j][i] - '0';
  }
  sort(a, a + m);
  for (int i = 0; i <= m; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % M;
  }
  b[0] = b[1] = 1;
  for (int i = 2; i <= m; i++) {
    for (int j = 0; j < i; j++) b[i] = (b[i] + c[i - 1][j] * b[j]) % M;
  }
  int ans = 1;
  for (int i = 0, j = 0; i < m; i = j) {
    for (; j < m && a[j] == a[i]; j++)
      ;
    int k = j - i;
    ans = ans * b[k] % M;
  }
  printf("%d\n", ans);
  return 0;
}
