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 v[1010];
long long f[1010][1010];
long long calc(int x) {
  memset(f, 0, sizeof(f));
  f[0][0] = 1;
  for (int i = 1; i <= x; i++)
    for (int j = 1; j <= i; j++)
      f[i][j] = (f[i - 1][j - 1] + f[i - 1][j] * j) % 1000000007;
  long long ret = 0;
  for (int i = 1; i <= x; i++) (ret += f[x][i]) %= 1000000007;
  return ret;
}
int main() {
  int m, n;
  scanf("%d %d", &m, &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      int val;
      scanf("%1d", &val);
      v[j] <<= 1;
      v[j] |= val;
    }
  sort(v + 1, v + m + 1);
  int cnt = 1;
  long long ans = 1;
  for (int i = 2; i <= m; i++)
    if (v[i] == v[i - 1])
      cnt++;
    else {
      (ans *= calc(cnt)) %= 1000000007;
      cnt = 1;
    }
  (ans *= calc(cnt)) %= 1000000007;
  printf("%I64d\n", ans);
  return 0;
}
