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 = 1011, p = 1000000007;
int C[maxn][maxn], f[maxn];
void preset() {
  int n = 1000;
  C[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    C[i][i] = 1;
    C[i][0] = 1;
    for (int j = 1; j < i; ++j) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % p;
  }
  f[0] = 1;
  f[1] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= i; ++j)
      f[i + 1] = (f[i + 1] + 1LL * f[j] * C[i][j]) % p;
}
int main() {
  preset();
  int m, n;
  static char s[maxn][maxn];
  scanf("%d%d", &m, &n);
  for (int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
  static map<string, int> mp;
  for (int i = 1; i <= m; ++i) {
    string S = "";
    for (int j = 1; j <= n; ++j) S += s[j][i];
    mp[S]++;
  }
  int ans = 1;
  for (auto x = mp.begin(); x != mp.end(); ++x) {
    ans = 1LL * ans * f[x->second] % p;
  }
  printf("%d\n", ans);
}
