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 m, n, ans = 1, si[1010][1010], sum[1010];
char str[1010][1010];
map<long long, long long> cnt;
map<long long, long long>::iterator it;
int main() {
  long long i, j, t;
  cin >> m >> n;
  for (i = 1; i <= n; i++) scanf("%s", str[i] + 1);
  si[0][0] = 1;
  for (i = 1; i <= m; i++) {
    for (j = 1; j <= i; j++) {
      si[i][j] = (si[i - 1][j - 1] + si[i - 1][j] * j) % 1000000007;
      sum[i] = (sum[i] + si[i][j]) % 1000000007;
    }
  }
  for (i = 1; i <= m; i++) {
    t = 0;
    for (j = 1; j <= n; j++) {
      t = t * 2 + str[j][i] - '0';
    }
    cnt[t]++;
  }
  for (it = cnt.begin(); it != cnt.end(); it++) {
    t = (*it).second;
    ans = ans * sum[t] % 1000000007;
  }
  cout << ans;
}
