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>
const int MAXN = 50 + 10, MAXM = 1000 + 10;
const int MOD = 1e9 + 7;
namespace solver1 {
long long mask[MAXM];
int choose[MAXM][MAXM], f[MAXM];
void main() {
  int n, m;
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; i++) {
    char buff[MAXM];
    scanf("%s", buff + 1);
    for (int j = 1; j <= m; j++) {
      mask[j] |= (1ll * (buff[j] - '0')) << i;
    }
  }
  std::sort(mask + 1, mask + m + 1);
  choose[0][0] = 1;
  for (int i = 1; i <= m; i++) {
    choose[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      choose[i][j] = choose[i - 1][j] + choose[i - 1][j - 1];
      choose[i][j] %= MOD;
    }
  }
  f[0] = 1;
  for (int i = 1; i <= m; i++) {
    f[i] = 1;
    for (int j = 1; j < i; j++) {
      f[i] += 1ll * f[j] * choose[i - 1][j] % MOD;
      f[i] %= MOD;
    }
  }
  int cnt = 1, ans = 1;
  for (int i = 2; i <= m; i++) {
    if (mask[i] == mask[i - 1]) {
      cnt++;
    } else {
      ans = 1ll * ans * f[cnt] % MOD;
      cnt = 1;
    }
  }
  ans = 1ll * ans * f[cnt] % MOD;
  printf("%d", ans);
}
}  // namespace solver1
int main() { solver1::main(); }
