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;
long long mask[(1LL << 20)];
long long p[(1LL << 20)];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long n;
  cin >> n;
  long long a[n];
  p[0] = 1;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    mask[a[i]]++;
    if (i != 0) p[i] = (p[i - 1] * 2) % 1000000007;
  }
  p[n] = (p[n - 1] * 2) % 1000000007;
  for (long long i = 0; i < 20; i++) {
    for (long long j = 0; j < (1LL << 20); j++) {
      if (!(j & (1 << i))) {
        mask[j] += mask[j ^ (1 << i)];
      }
    }
  }
  for (long long i = 0; i < (1LL << 20); i++) {
    mask[i] = (p[mask[i]] - 1 + 1000000007) % 1000000007;
  }
  for (long long i = 0; i < 20; i++) {
    for (long long j = 0; j < (1LL << 20); j++) {
      if (!(j & (1LL << i))) {
        mask[j] = (mask[j] - mask[j ^ (1LL << i)] + 1000000007) % 1000000007;
      }
    }
  }
  cout << mask[0] << " ";
  return 0;
}