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;
int fastMax(int x, int y) { return (((y - x) >> (32 - 1)) & (x ^ y)) ^ y; }
int fastMin(int x, int y) { return (((y - x) >> (32 - 1)) & (x ^ y)) ^ x; }
const long long int N = 20;
const long long int MAXN = 1 << N;
const long long int MOD = 1e9 + 7;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    x = (x * x) % MOD;
    y >>= 1;
  }
  return res;
}
long long int A[MAXN], F[MAXN];
void solve() {
  long long int n;
  cin >> n;
  for (int i = (1); i <= (n); ++i) {
    cin >> A[i];
    F[A[i]]++;
  }
  for (int i = (0); i <= (N - 1); ++i) {
    for (int mask = (0); mask <= (MAXN - 1); ++mask) {
      if (mask & (1 << i)) F[mask ^ (1 << i)] += F[mask];
    }
  }
  long long int ans = 0;
  for (int i = (0); i <= (MAXN - 1); ++i) {
    long long int b = __builtin_popcount(i);
    long long int cnt = power(2, F[i]);
    if (b & 1) cnt = MOD - cnt;
    ans += cnt;
    ans %= MOD;
  }
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  t = 1;
  while (t--) solve();
}