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;
using ll = long long;
using vi = vector<int>;
using ii = pair<int, int>;
const int oo = 0x3f3f3f3f;
const ll OO = ll(oo) * oo;
const int N = 1 << 20;
const ll M = 1000000007;
ll A[N];
ll f[N];
ll p2[N];
void add(ll& x, ll y) {
  x += y;
  if (x >= M) x -= M;
  if (x < 0) x += M;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    f[A[i]]++;
  }
  p2[0] = 1;
  for (int i = 1; i < N; i++) p2[i] = p2[i - 1] * 2 % M;
  for (int b = 1; b < N; b <<= 1) {
    for (int u = 0; u < N; u += (b << 1)) {
      for (int l = 0; l < b; l++) {
        add(f[u | l], f[u | l | b]);
      }
    }
  }
  for (int i = 0; i < N; i++) {
    f[i] = p2[f[i]] - 1;
  }
  for (int b = 1; b < N; b <<= 1) {
    for (int u = 0; u < N; u += (b << 1)) {
      for (int l = 0; l < b; l++) {
        add(f[u | l], -f[u | l | b]);
      }
    }
  }
  cout << f[0] << '\n';
  return 0;
}