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>
const double eps = 0.000000001;
using namespace std;
const int mx_n = 1024 * 1024;
int a[mx_n];
int F[mx_n];
int two[mx_n];
void solve() {
  int N = 20;
  int n;
  cin >> n;
  fill(F, F + mx_n, 0);
  fill(a, a + mx_n, 0);
  vector<int> b;
  int x;
  for (int i = 0; i < n; i++) {
    cin >> x;
    a[((1 << N) - 1) ^ x]++;
  }
  two[0] = 1;
  for (int i = 1; i < mx_n; i++) {
    two[i] = two[i - 1] * 2;
    if (two[i] > 1000000007) two[i] -= 1000000007;
  }
  for (int i = 0; i < (1 << N); ++i) F[i] = a[i];
  for (int i = 0; i < N; ++i)
    for (int mask = 0; mask < (1 << N); ++mask) {
      if (mask & (1 << i)) F[mask] += F[mask ^ (1 << i)];
    }
  long long ans = 0;
  for (int i = 0; i < (1 << N); i++) {
    int tmp = two[F[((1 << N) - 1) ^ i]];
    int u = __builtin_popcount(i);
    if (u % 2 == 1) tmp *= -1;
    ans = (tmp + ans) % 1000000007;
  }
  ans = (ans + 1000000007) % 1000000007;
  printf("%lld", ans);
}
int main(int argc, char const *argv[]) {
  int t = 1;
  while (t--) {
    solve();
  }
}