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 maxN = 1004;
const int MOD = 1e9 + 7;
int n, m;
char str[maxN];
long long mask[maxN];
int c[maxN][maxN];
int bell[maxN];
map<long long, int> Map;
int main() {
  scanf("%d %d\n", &m, &n);
  for (int i = 1; i <= n; i++) {
    gets(str + 1);
    for (int j = 1; j <= m; j++)
      if (str[j] == '1') mask[j] |= (1LL << (i - 1));
  }
  for (int i = 1; i <= m; i++) Map[mask[i]]++;
  for (int i = 0; i <= m; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;
  }
  bell[0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < i; j++)
      bell[i] = (bell[i] + 1LL * c[i - 1][j] * bell[j] % MOD) % MOD;
  int sol = 1;
  for (map<long long, int>::iterator it = Map.begin(); it != Map.end(); it++)
    sol = (1LL * sol * bell[it->second]) % MOD;
  cout << sol;
  return 0;
}
