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 mod = 1e9 + 7;
int c[1010][1010], b[1010];
char s[66][1010];
long long a[1010];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i <= n; i++) c[0][i] = 0, c[i][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      if (c[i][j] >= mod) c[i][j] -= mod;
    }
  b[0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < i; j++) b[i] = (b[i] + 1ll * c[i - 1][j] * b[j]) % mod;
  for (int i = 0; i < m; i++) scanf("%s", s[i] + 1);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < m; j++)
      if (s[j][i] == '1') a[i] ^= 1ll << j;
  sort(a + 1, a + n + 1);
  int ans = 1;
  for (int i = 1, j = 1; i <= n; i = j) {
    while (j <= n && a[i] == a[j]) j++;
    ans = 1ll * ans * b[j - i] % mod;
  }
  cout << ans << endl;
}
