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 N = 1e3 + 10;
const int MOD = 1e9 + 7;
long long C[N][N];
map<long long, int> mp;
long long a[N], b[N], c[N][N];
int main() {
  for (int i = 0; i < N; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
    }
  }
  int n, m;
  cin >> m >> n;
  while (n--) {
    long long y = 0;
    for (int i = 0; i < m; i++) {
      long long x;
      scanf("%1lld", &x);
      a[i] |= x << n;
    }
  }
  for (int i = 0; i < m; i++) {
    mp[a[i]]++;
  }
  b[0] = b[1] = 1;
  for (int i = 2; i <= m; i++) {
    for (int j = 0; j < i; j++) {
      (b[i] += C[i - 1][j] * b[i - j - 1]) %= MOD;
    }
  }
  long long ans = 1;
  for (auto it : mp) {
    (ans *= b[it.second]) %= MOD;
  }
  cout << ans << endl;
  return 0;
}
