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 N = 1005;
const int mod = 1e9 + 7;
map<long long, int> Map;
int n, m, x, y;
long long a[N], bel[N], dp[N][N], ans;
char s[N];
int main(void) {
  scanf("%d%d", &m, &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", s + 1);
    for (int j = 1; j <= m; ++j) a[j] = a[j] * 2 + s[j] - '0';
  }
  for (int i = 1; i <= m; ++i) ++Map[a[i]];
  bel[0] = 1;
  for (int i = 1; i <= m; ++i) {
    dp[i][1] = 1;
    for (int j = 2; j <= i; ++j) {
      dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j] * j) % mod;
    }
    for (int j = 1; j <= i; ++j) bel[i] = (dp[i][j] + bel[i]) % mod;
  }
  ans = 1;
  for (map<long long, int>::iterator it = Map.begin(); it != Map.end(); ++it)
    ans = ans * bel[(it->second)] % mod;
  cout << ans;
}
