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;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
const long long N = 1000 + 7;
long long test;
long long m, n;
string t[N];
long long res;
long long dp[N][N];
long long final[N];
map<string, long long> mm;
void solve() {
  cin >> m >> n;
  for (long long i = 1; i <= (n); ++i) {
    cin >> t[i];
  }
  dp[1][1] = 1;
  for (long long i = 2; i <= (m); ++i) {
    for (long long j = 1; j <= (m); ++j) {
      dp[i][j] = (dp[i - 1][j] * j + dp[i - 1][j - 1]) % 1000000007;
    }
  }
  for (long long i = 1; i <= (m); ++i) {
    for (long long j = 1; j <= (m); ++j) {
      final[i] = (final[i] + dp[i][j]) % 1000000007;
    }
  }
  for (long long i = 0; i < (m); ++i) {
    string s;
    s.clear();
    for (long long j = 1; j <= (n); ++j) {
      s += t[j][i];
    }
    mm[s]++;
  }
  res = 1;
  for (auto it : mm) {
    res = res * final[it.second] % 1000000007;
  }
  cout << res;
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  cout << fixed;
  test = 1;
  while (test--) {
    solve();
  }
  return 0;
}
