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 ll = long long;
using ld = long double;
using namespace std;
const int MOD = 1000 * 1000 * 1000 + 7;
ll pw(ll a, ll b = MOD - 2) {
  if (b == 0) {
    return 1;
  }
  ll v = pw(a, b / 2);
  v = (v * v) % MOD;
  if (b & 1) {
    v = (v * a) % MOD;
  }
  return v;
}
constexpr int MAXN = 1001;
ll f[MAXN];
ll a[MAXN];
ll cnk[MAXN][MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cout.setf(ios::fixed);
  cout.precision(20);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int m, n;
  cin >> m >> n;
  cnk[0][0] = 1;
  for (int i = 1; i <= m; ++i) {
    cnk[i][0] = 1;
    for (int j = 1; j <= i; ++j) {
      cnk[i][j] = (cnk[i - 1][j] + cnk[i - 1][j - 1]) % MOD;
    }
  }
  f[0] = 1;
  for (int i = 1; i <= m; ++i) {
    for (int j = 0; j < i; ++j) {
      f[i] = (f[i] + cnk[i - 1][j] * f[i - j - 1]) % MOD;
    }
  }
  for (int i = 0; i < n; ++i) {
    ll val = (1ll << i);
    string s;
    cin >> s;
    for (int j = 0; j < m; ++j) {
      a[j] += (s[j] - '1') * val;
    }
  }
  ll ans = 1;
  sort(a, a + m);
  for (int l = 0, r = 0; l < m; l = r) {
    while (r < m && a[l] == a[r]) {
      ++r;
    }
    ans = (ans * f[r - l]) % MOD;
  }
  cout << ans << "\n";
}
