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 double pi = acos(-1.0);
const int N = 20;
const int M = 1000005;
int n, ans = 0;
int F[M], a[M];
void calculateF() {
  for (int i = 0; i < N; i += 1) {
    for (int mask = M; mask >= 0; mask -= 1) {
      if (mask & (1 << i)) {
        (F[1 << i ^ mask] += F[mask]) %= 1000000007;
      }
    }
  }
  for (int i = M; i >= 0; i -= 1) {
    (ans += 1ll * (__builtin_popcount(i) % 2 ? -1 : 1) * (a[F[i]] - 1)) %=
        1000000007;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  memset(F, (0), sizeof(F));
  memset(a, (0), sizeof(a));
  for (int i = 0; i < n; i += 1) {
    int x;
    cin >> x;
    F[x]++;
  }
  a[0] = 1;
  for (int i = 0; i < n + 1; i += 1) {
    if (i == 0) continue;
    a[i] = (a[i - 1] << 1) % 1000000007;
  }
  calculateF();
  cout << ((ans + 1000000007) % 1000000007);
  return 0;
}