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 N = 1005, M = 1e9 + 7;
long long n, m, C[N][N], B[N], ans, a[N];
map<long long, long long> mp;
char s[N];
signed main() {
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= m; i++) {
    scanf("%s", s + 1);
    for (long long j = 1; j <= n; j++) {
      long long t = s[j] == '1';
      a[j] |= t << i - 1;
    }
  }
  for (long long i = 1; i <= n; i++) mp[a[i]]++;
  for (long long i = 0; i <= n; i++) C[i][0] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % M;
  B[0] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = 0; j < i; j++)
      B[i] = (B[i] + C[i - 1][j] * B[j] % M) % M;
  ans = 1;
  for (auto x : mp) (ans *= B[x.second] % M) %= M;
  printf("%lld", ans);
  return 0;
}
