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 MOD = 1000000007;
long long binom[1001][1001];
long long bell[1001];
int main() {
  binom[0][0] = 1;
  bell[0] = 1;
  for (int i = 1; i <= 1000; i++) {
    binom[i][0] = 1;
    for (int j = 1; j <= 1000; j++) {
      binom[i][j] = binom[i - 1][j - 1] + binom[i - 1][j];
      if (binom[i][j] >= MOD) binom[i][j] -= MOD;
      bell[i] += binom[i - 1][j - 1] * bell[j - 1] % MOD;
      if (bell[i] >= MOD) bell[i] -= MOD;
    }
  }
  int n, m;
  cin >> m >> n;
  bitset<1000> full;
  for (int i = 0; i < m; i++) full[i] = true;
  bitset<1000> masks[n];
  for (int i = 0; i < n; i++) {
    string str;
    cin >> str;
    for (int j = 0; j < m; j++) masks[i][j] = str[j] == '1';
  }
  bitset<1000> used;
  long long ans = 1;
  for (int j1 = 0; j1 < m; j1++) {
    if (used[j1]) continue;
    bitset<1000> equiv;
    for (int j2 = 0; j2 < m; j2++) {
      bool ok = true;
      for (int i = 0; i < n && ok; i++)
        if (masks[i][j1] != masks[i][j2]) ok = false;
      if (ok) equiv[j2] = true;
    }
    if (equiv.count() == 0)
      ans *= 2;
    else
      ans *= bell[equiv.count()];
    ans %= MOD;
    used |= equiv;
  }
  cout << ans << endl;
}
