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 C[1010][1010];
long long bell[1010];
char buf[1010];
string input[60];
map<string, int> mp;
int main(void) {
  int K, N, i, j;
  cin >> N >> K;
  for ((i) = 0; (i) < (int)(K); (i)++) {
    scanf("%s", buf);
    input[i] = buf;
  }
  for ((i) = 0; (i) < (int)(N); (i)++) {
    string s;
    for ((j) = 0; (j) < (int)(K); (j)++) s += input[j][i];
    mp[s]++;
  }
  for ((i) = 0; (i) < (int)(N + 1); (i)++)
    for ((j) = 0; (j) < (int)(i + 1); (j)++) {
      if (j == 0 || j == i)
        C[i][j] = 1;
      else
        C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 1000000007ll;
    }
  bell[0] = 1;
  for (i = 1; i <= N; i++)
    for (j = 1; j <= i; j++)
      bell[i] = (bell[i] + bell[i - j] * C[i - 1][j - 1]) % 1000000007ll;
  long long ans = 1;
  for (__typeof((mp).begin()) itr = (mp).begin(); itr != (mp).end(); itr++) {
    int c = (itr->second);
    ans = ans * bell[c] % 1000000007ll;
  }
  cout << ans << endl;
  return 0;
}
