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 MAXN = 1005, MOD = 1000000007;
int f[MAXN][MAXN], g[MAXN];
long long a[MAXN];
char b[55][MAXN];
int main() {
  int m, n;
  scanf("%d%d", &m, &n);
  f[0][0] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= i; j++)
      f[i][j] = (f[i - 1][j - 1] + (long long)f[i - 1][j] * j) % MOD;
    for (int j = 1; j <= i; j++) g[i] = (g[i] + f[i][j]) % MOD;
  }
  for (int i = 0; i < n; i++) scanf("%s", b[i]);
  for (int i = 0; i < m; i++)
    for (int j = 0; j < n; j++)
      if (b[j][i] == '1') a[i] |= 1LL << j;
  sort(a, a + m);
  int ans = 1;
  for (int i = 0; i < m;) {
    int j;
    for (j = i + 1; j < m && a[j] == a[i]; j++)
      ;
    ans = (long long)ans * g[j - i] % MOD;
    i = j;
  }
  printf("%d", ans);
  return 0;
}
