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;
struct Modint {
  int val;
  Modint(int new_val = 0) { val = new_val; }
  Modint operator+(Modint r) { return (val + r.val) % MOD; }
  Modint operator-(Modint r) { return (val + MOD - r.val) % MOD; }
  Modint operator*(Modint r) { return 1LL * val * r.val % MOD; }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> m >> n;
  vector<string> ts(n);
  for (int i = 0; i < n; i++) {
    cin >> ts[i];
  }
  vector<long long> descriptors;
  for (int j = 0; j < m; j++) {
    long long val = 0;
    for (int i = 0; i < n; i++) {
      val = val * 2 + (ts[i][j] == '1');
    }
    descriptors.push_back(val);
  }
  sort(descriptors.begin(), descriptors.end());
  static Modint dp[1001][1001];
  dp[0][0] = 1;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j <= i; j++) {
      dp[i + 1][j] = dp[i + 1][j] + dp[i][j] * j;
      dp[i + 1][j + 1] = dp[i + 1][j + 1] + dp[i][j];
    }
  }
  Modint result = 1;
  for (int i = 0, j = 0; i < m; i = j) {
    while (j < m && descriptors[i] == descriptors[j]) {
      j++;
    }
    Modint to_mult = 0;
    for (int k = 0; k <= j - i; k++) {
      to_mult = to_mult + dp[j - i][k];
    }
    result = result * to_mult;
  }
  cout << result.val << '\n';
  return 0;
}
