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 = 1010, MOD = 1e9 + 7;
long long c[maxn][maxn], f[maxn], n, m;
long long b[maxn];
map<long long, long long> mp;
int main() {
  scanf("%lld%lld", &m, &n);
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= m; j++) {
      int x;
      scanf("%1d", &x);
      b[j] += (1ll * x) << i;
    }
  }
  for (int i = 1; i <= m; i++) mp[b[i]]++;
  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]) % MOD;
  }
  f[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j < i; j++)
      f[i] = (f[i] + 1ll * c[i - 1][j] * f[j] % MOD) % MOD;
  }
  long long ans = 1;
  for (map<long long, long long>::iterator it = mp.begin(); it != mp.end();
       it++)
    ans = 1ll * ans * f[it->second] % MOD;
  printf("%lld", ans);
  return 0;
}
