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>
long long memo[(1 << 21)];
long long modexp(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = ((res % 1000000007) * (a % 1000000007)) % 1000000007;
    a = ((a % 1000000007) * (a % 1000000007)) % 1000000007;
    n >>= 1;
  }
  return res;
}
int main(void) {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  int n;
  std::cin >> n;
  for (int i = 0; i < n; i++) {
    int t;
    std::cin >> t;
    memo[t]++;
  }
  for (int p = 0; p < 21; p++) {
    for (int mask = ((1 << 21) - 1); mask >= 0; mask--) {
      if (!(mask & (1 << p))) memo[mask] += memo[mask ^ (1 << p)];
    }
  }
  long long ans = 0;
  for (int mask = 0; mask < (1 << 21); mask++) {
    int z = __builtin_popcount(mask);
    if (z & 1)
      ans = ((ans % 1000000007) - ((modexp(2, memo[mask]) - 1) % 1000000007) +
             1000000007) %
            1000000007;
    else
      ans = ((ans % 1000000007) + ((modexp(2, memo[mask]) - 1) % 1000000007)) %
            1000000007;
  }
  std::cout << ans << "\n";
  return 0;
}