Problem: Jzzhu have n non-negative integers a1, a2, ..., an. We will call a sequence of indexes i1, i2, ..., ik (1 ≤ i1 < i2 < ... < ik ≤ n) a group of size k. 

Jzzhu wonders, how many groups exists such that ai1 & ai2 & ... & aik = 0 (1 ≤ k ≤ n)? Help him and print this number modulo 1000000007 (109 + 7). Operation x & y denotes bitwise AND operation of two numbers.

Input

The first line contains a single integer n (1 ≤ n ≤ 106). The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 106).

Output

Output a single integer representing the number of required groups modulo 1000000007 (109 + 7).

Examples

Input

3
2 3 3


Output

0


Input

4
0 1 2 3


Output

10


Input

6
5 2 0 5 2 1


Output

53

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int m = 20;
const int mod = 1e9 + 7;
const int maxn = (1 << m);
long long d[maxn][m + 1];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, a;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    d[a][0]++;
    d[a][0] *= 2;
    d[a][0]--;
    d[a][0] %= mod;
  }
  for (int mask = (1 << m) - 1; mask >= 0; mask--) {
    for (int i = 1; i <= m; i++) {
      if (!((1 << (i - 1)) & mask))
        d[mask][i] =
            (d[mask][i - 1] + 1) * (d[mask ^ (1 << (i - 1))][i - 1] + 1) - 1;
      else
        d[mask][i] += d[mask][i - 1];
      d[mask][i] %= mod;
    }
  }
  long long ans = 0;
  for (int i = 0; i < (1 << m); i++) {
    if (__builtin_popcount(i) % 2 == 0)
      ans += d[i][m];
    else
      ans -= d[i][m];
    ans += mod;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}