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 n, m, b[1005][1005], ans, c[1005];
long long a[1005];
char s[1005];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", s + 1);
    for (int j = 1; j <= n; j++) a[j] = a[j] * 2 | (s[j] == '1');
  }
  sort(a + 1, a + n + 1);
  b[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      b[i][j] = (b[i - 1][j - 1] + 1ll * b[i - 1][j] * j) % 1000000007;
      c[i] = (c[i] + b[i][j]) % 1000000007;
    }
  }
  int now = 0, ans = 1;
  for (int i = 1; i <= n; i++) {
    now++;
    if (i == n || a[i] != a[i + 1]) {
      ans = 1ll * ans * c[now] % 1000000007;
      now = 0;
    }
  }
  printf("%d\n", ans);
}
