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 mod = 1e9 + 7;
const int N = 1050;
int dp[N][N], sz[N], ans = 1;
long long a[N];
char ch[N];
map<long long, int> Map;
map<long long, int>::iterator it;
int main() {
  int n, m, i, j;
  dp[0][0] = 1;
  scanf("%i %i", &m, &n);
  for (i = 1; i <= m; i++)
    for (j = 1; j <= i; j++) {
      dp[i][j] = (long long)dp[i - 1][j] * j % mod + dp[i - 1][j - 1];
      dp[i][j] %= mod;
      sz[i] += dp[i][j];
      sz[i] %= mod;
    }
  for (i = 0; i < n; i++) {
    scanf("%s", &ch);
    for (j = 0; j < m; j++) a[j] += ((long long)(ch[j] - '0')) << i;
  }
  for (j = 0; j < m; j++) Map[a[j]]++;
  for (it = Map.begin(); it != Map.end(); it++) {
    ans = (long long)ans * sz[(it->second)] % mod;
  }
  printf("%i\n", ans);
  return 0;
}
