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;
  long long masks[m];
  memset(masks, 0, sizeof masks);
  for (int i = 0; i < n; i++) {
    string str;
    cin >> str;
    for (int j = 0; j < m; j++)
      if (str[j] == '1') masks[j] |= 1LL << i;
  }
  sort(masks, masks + m);
  int cur = 1;
  long long ans = 1;
  for (int i = 1; i < m; i++)
    if (masks[i] == masks[i - 1])
      cur++;
    else {
      ans = ans * bell[cur] % MOD;
      cur = 1;
    }
  ans = ans * bell[cur] % MOD;
  cout << ans << endl;
}
