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, ans = 1, bell[1005], c[1005][1005];
char s[55][1005];
map<long long, int> w;
long long get(int x) {
  long long res = 0;
  for (int i = 1; i <= n; i++) res = (res << 1) + s[i][x] - 48;
  return res;
}
int main() {
  scanf("%d%d", &m, &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) scanf(" %c", &s[i][j]);
  for (int i = 1; i <= m; i++) w[get(i)]++;
  c[0][0] = 1;
  bell[0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 0; j <= i; j++) {
      c[i][j] = (c[i - 1][j - 1] + (long long)j * c[i - 1][j]) % 1000000007;
      bell[i] = (bell[i] + c[i][j]) % 1000000007;
    }
  for (int i = 1; i <= m; i++) {
    long long num = get(i);
    ans = (long long)ans * bell[w[num]] % 1000000007;
    w[num] = 0;
  }
  printf("%d\n", ans);
  return 0;
}
