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 = (int)1e9 + 7;
int add(int a, int b) {
  long long c = (long long)a + b;
  while (c < 0) c += mod;
  return c % mod;
}
int mult(int a, int b) {
  long long c = (long long)a * b;
  return c % mod;
}
const int MAXN = 1010;
int Bell[MAXN];
void computeBell() {
  vector<int> v;
  v.push_back(1);
  for (int i = 0; i < MAXN; i++) {
    Bell[i] = v[0];
    vector<int> vi;
    vi.push_back(v.back());
    for (int j : v) {
      vi.push_back(add(vi.back(), j));
    }
    v = vi;
  }
}
int m, n;
vector<string> v(MAXN, "");
char s[MAXN];
void load() {
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    for (int j = 0; j < m; j++) v[j] = v[j] + s[j];
  }
}
void solve() {
  computeBell();
  int sol = 1;
  sort(v.begin(), v.begin() + m);
  string cur = "";
  int br = 1;
  for (string s : v) {
    if (s == cur) {
      ++br;
    } else {
      sol = mult(sol, Bell[br]);
      cur = s;
      br = 1;
    }
  }
  printf("%d\n", sol);
}
int main() {
  load();
  solve();
  return 0;
}
