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 N = 1002;
const int M = 1000000007;
long long c[N][N], b[N], ans = 1;
void init() {
  for (int i = 0; i < N; i++) c[i][0] = 1;
  for (int i = 1; i < N; i++) {
    for (int j = 1; j <= i; j++) {
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % M;
    }
  }
  b[0] = b[1] = 1;
  for (int i = 2; i < N; i++) {
    for (int j = 0; j < i; j++) {
      b[i] = (b[i] + c[i - 1][j] * b[j] % M) % M;
    }
  }
}
int n, m;
char s[52][1002];
vector<int> v[1002];
int main() {
  init();
  scanf("%d %d", &m, &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s[i] + 1);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      v[j].push_back(s[i][j]);
    }
  }
  sort(v + 1, v + 1 + m);
  int cur = 1;
  for (int i = 2; i <= m; i++) {
    if (v[i] == v[i - 1]) cur++;
    if (v[i] != v[i - 1] || i == m) {
      ans = ans * b[cur] % M;
      cur = 1;
    }
  }
  printf("%lld\n", ans);
}
