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;
int mod = 1000000007;
string a[100];
bool used[100100];
long long fact[100100];
long long ifact[100100];
long long d[100100];
long long pw(long long a, long long n) {
  long long res = 1, sq = a % mod;
  while (n) {
    if (n & 1) {
      res *= sq;
      res %= mod;
    }
    sq = (sq * sq) % mod;
    n /= 2;
  }
  return res;
}
long long inv(long long x) { return pw(x, mod - 2); }
int main() {
  fact[0] = 1;
  for (int i = 0; i < (int)(100099); ++i) {
    fact[i + 1] = fact[i] * (i + 1) % mod;
  }
  for (int i = 0; i < (int)(100100); ++i) {
    ifact[i] = inv(fact[i]);
  }
  d[0] = 1;
  for (int i = 1; i < 1010; ++i) {
    for (int j = 0; j < (int)(i); ++j) {
      long long cur = fact[i - 1] * ifact[j] % mod;
      cur *= ifact[i - 1 - j];
      cur %= mod;
      cur *= d[j];
      d[i] += cur % mod;
      d[i] %= mod;
    }
  }
  ios::sync_with_stdio(false);
  int n, m;
  cin >> m >> n;
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i];
  }
  long long ans = 1;
  for (int j = 0; j < (int)(m); ++j) {
    if (used[j]) {
      continue;
    }
    int x = 1;
    for (int k = j + 1; k < m; ++k) {
      bool ok = true;
      for (int i = 0; i < (int)(n); ++i) {
        if (a[i][j] != a[i][k]) {
          ok = false;
        }
      }
      if (ok) {
        ++x;
        used[k] = true;
      }
    }
    ans *= d[x];
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}
