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;
const long long inf = 1e9;
const long long N = (1 << 20);
long long mod_expo(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & (long long)1) ans = (ans * a) % c;
    a = (a * a) % c;
    b = b >> 1;
  }
  return ans;
}
long long inverse(long long n) { return mod_expo(n, mod - 2, mod); }
int n;
int total;
int a[N];
long long f[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  total = 1 << 20;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a[x]++;
  }
  n = 20;
  for (int i = 0; i < total; i++) f[i] = a[i];
  for (int i = 0; i < n; i++) {
    for (int mask = total - 1; mask >= 0; mask--) {
      if ((mask & (1 << i)) == 0) f[mask] += f[mask ^ (1 << i)];
    }
  }
  long long ans = 0;
  for (int i = 0; i < total; i++) {
    int bits = __builtin_popcount(i);
    if (bits % 2)
      ans -= mod_expo(2, f[i], mod);
    else
      ans += mod_expo(2, f[i], mod);
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  cout << ans;
  return 0;
}