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 MX = 1005, mod = 1000000007;
int n, m;
bitset<MX> bs, s[MX];
long long res = 1, comb[MX][MX], dp[MX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 0; i < MX; i++) {
    comb[i][0] = 1;
    for (int j = 1; j <= i; j++)
      comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % mod;
  }
  dp[0] = dp[1] = 1;
  for (int i = 2; i < MX; i++)
    for (int j = 0; j < i; j++)
      (dp[i] += dp[i - j - 1] * comb[i - 1][j]) %= mod;
  cin >> m >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      char c;
      cin >> c;
      s[i][j] = c == '1';
    }
  }
  for (int i = 0; i < m; i++)
    if (!bs[i]) {
      bitset<MX> x;
      for (int j = 0; j < m; j++) x[j] = 1;
      for (int j = 0; j < n; j++) {
        if (s[j][i])
          x &= s[j];
        else
          x &= ~s[j];
      }
      (res *= dp[x.count()]) %= mod;
      bs |= x;
    }
  cout << res << '\n';
  return 0;
}
