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;
long long m, n;
char a[55][1005];
long long dp[1005][1005];
map<string, long long> mp;
vector<string> save;
int main() {
  scanf("%I64d%I64d", &m, &n);
  dp[0][0] = 1;
  for (long long i = 1; i <= 1001; i++)
    for (long long j = 0; j <= i; j++) {
      if (j == 0)
        dp[i][j] = dp[i - 1][i - 1];
      else
        dp[i][j] = (dp[i][j - 1] + dp[i - 1][j - 1]) % 1000000007;
    }
  for (long long i = 0; i < n; i++) scanf("%s", a[i]);
  save.clear();
  for (long long i = 0; i < m; i++) {
    string s = "";
    for (long long j = 0; j < n; j++) s += a[j][i];
    if (mp.find(s) == mp.end()) {
      mp[s] = 1;
      save.push_back(s);
    } else
      mp[s]++;
  }
  long long ans = 1;
  for (long long i = 0; i < save.size(); i++) {
    string s = save[i];
    ans = (ans * dp[mp[s]][0]) % 1000000007;
  }
  printf("%I64d\n", ans);
  return 0;
}
