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, c[1010][1010], f[1010], b[1010], x, ans = 1;
const long long mo = (long long)1e9 + 7;
map<long long, long long> mp;
int main() {
  for (cin >> m >> n; n--;)
    for (int i = 0; i < m; ++i) scanf("%1lld", &x), b[i] |= x << n;
  for (int i = 0; i < m; ++i) ++mp[b[i]];
  for (int i = 0; i <= m; ++i)
    for (int j = 0; j <= i; ++j)
      c[i][j] = (j > 0) ? (c[i - 1][j - 1] + c[i - 1][j]) % mo : 1;
  f[0] = f[1] = 1;
  for (int i = 2; i <= m; ++i)
    for (int j = 0; j < i; ++j) f[i] = (f[i] + c[i - 1][j] * f[j]) % mo;
  for (auto o : mp) ans = (ans * f[o.second]) % mo;
  cout << ans << endl;
}
