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 long long mod = 1e9 + 7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, bit;
  cin >> n;
  vector<int> a(n);
  vector<long long> p(n + 1);
  p[0] = 1;
  int mx = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mx = max(mx, a[i]);
    p[i + 1] = p[i] * 2 % mod;
  }
  bit = log2(mx) + 1;
  vector<int> f(1 << bit);
  for (auto x : a) {
    f[x]++;
  }
  for (int i = 0; i < bit; i++) {
    for (int mask = 0; mask < 1 << bit; mask++) {
      if (mask >> i & 1) {
        f[mask ^ (1 << i)] += f[mask];
      }
    }
  }
  long long ans = p[n] - 1;
  for (int i = 1; i <= mx; i++) {
    if (__builtin_popcount(i) & 1) {
      ans = (ans - p[f[i]] + 1) % mod;
    } else {
      ans = (ans + p[f[i]] - 1) % mod;
    }
  }
  cout << (ans + mod) % mod << endl;
  return 0;
}