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;
long long A[1100];
int B[1100], father[1100];
int C[1100][1100];
const int mo = 1e9 + 7;
int n, m;
char ch[1100];
int quick(int k1, int k2) {
  int k3 = 1;
  while (k2) {
    if (k2 & 1) k3 = 1ll * k3 * k1 % mo;
    k2 >>= 1;
    k1 = 1ll * k1 * k1 % mo;
  }
  return k3;
}
int findfather(int k1) {
  if (father[k1] == k1) return k1;
  return father[k1] = findfather(father[k1]);
}
int main() {
  scanf("%d%d", &m, &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", ch + 1);
    for (int j = 1; j <= m; j++)
      if (ch[j] == '1') A[j] |= (1ll << i - 1);
  }
  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]) % mo;
  }
  B[0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= i; j++)
      B[i] = (B[i] + 1ll * B[i - j] * C[i - 1][j - 1]) % mo;
  for (int i = 1; i <= m; i++) father[i] = i;
  for (int i = 1; i <= m; i++)
    for (int j = i + 1; j <= m; j++)
      if (A[j] == A[i]) father[i] = findfather(father[j]);
  int ans = 1;
  for (int i = 1; i <= m; i++)
    if (father[i] == i) {
      int num = 0;
      for (int j = 1; j <= m; j++)
        if (findfather(father[j]) == i) num++;
      ans = 1ll * ans * B[num] % mo;
    }
  cout << ans << endl;
}
