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 long long P = 1000000007;
bool ban[1010][1010];
int bel[1010];
int n, m;
long long ans;
long long tag[1010], f[1010][1010], s[1010];
char str[1010];
int main() {
  scanf("%d%d", &m, &n);
  int i, j, siz;
  for (i = 1; i <= n; i++) {
    scanf("%s", str + 1);
    for (j = 1; j <= m; j++)
      if (str[j] == '1') tag[j] |= 1ll << (i - 1);
  }
  ans = f[0][0] = 1;
  for (i = 1; i <= m; i++)
    for (j = 1; j <= i; j++)
      f[i][j] = (f[i - 1][j] * j + f[i - 1][j - 1]) % P,
      s[i] = (s[i] + f[i][j]) % P;
  for (i = 1; i <= m; i++)
    if (!bel[i]) {
      siz = 0;
      for (j = i; j <= m; j++)
        if (tag[j] == tag[i]) bel[j] = i, siz++;
      ans = ans * s[siz] % P;
    }
  printf("%lld", ans);
  return 0;
}
