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 int n;
long long int first[(1LL << 20) + 2];
long long int expo(long long int base, long long int exponent,
                   long long int mod) {
  long long int ans = 1;
  while (exponent != 0) {
    if ((exponent & 1) == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base = base * base;
    base %= mod;
    exponent >>= 1;
  }
  return ans % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int i, j;
  cin >> n;
  memset(first, 0, sizeof first);
  for (i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    first[x]++;
  }
  for (long long int i = 0; i < 20; ++i) {
    for (long long int mask = 0; mask < (1LL << 20); ++mask) {
      if (!(mask & (1LL << i))) {
        first[mask] += first[mask | (1LL << i)];
      }
    }
  }
  long long int ans = 0;
  for (i = 0; i < (1LL << 20); i++) {
    long long int setbits = (long long int)(__builtin_popcount(i));
    if (setbits % 2)
      ans = (ans - expo(2LL, first[i], 1000000007) + 1000000007) % 1000000007;
    else
      ans = (ans + expo(2LL, first[i], 1000000007)) % 1000000007;
  }
  cout << ans;
  return 0;
}