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 int dp[10011] = {0};
bool vis[10001] = {0};
long long int fac[2011];
const long long int mod = (long long int)1e9 + 7;
long long inv(long long a, long long b = mod) {
  return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
long long int c(long long int n, long long int k) {
  assert(n >= k);
  long long int ans = fac[n];
  ans *= inv(fac[k]);
  ans %= mod;
  ans *= inv(fac[n - k]);
  ans %= mod;
  return ans;
}
long long int pre(long long int count) {
  if (count == 0) return 1;
  if (vis[count]) return dp[count];
  long long int &ans = dp[count];
  vis[count] = true;
  for (long long int i = 1; i <= count; i++) {
    ans += pre(count - i) * c(count - 1, i - 1);
    ans %= mod;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  fac[0] = 1;
  for (long long int i = 1; i <= (long long int)(2000); i++)
    fac[i] = fac[i - 1] * i, fac[i] %= mod;
  pre(1001);
  long long int m, n;
  cin >> m >> n;
  string s;
  vector<long long int> bits[1000];
  for (long long int i = 0; i < (long long int)(n); i++) {
    cin >> s;
    for (long long int j = 0; j < (long long int)(m); j++) {
      if (s[j] == '1') bits[j].push_back(i);
    }
  }
  map<vector<long long int>, long long int> sz;
  for (long long int i = 0; i < (long long int)(m); i++) {
    sz[bits[i]]++;
  }
  long long int ans = 1;
  for (auto p : sz) {
    ans *= dp[p.second];
    ans %= mod;
  }
  cout << ans;
}
