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 int block = 400;
const long long base = 31, base2 = 37;
const long long mod = 1e9 + 7, mod2 = 1e9 + 9;
int a[1000011], n, freq[(1 << 22) + 10];
long long f[(1 << 22) + 10];
long long quickPow(int x, int y) {
  if (y == 0) return 1;
  long long tmp = quickPow(x, y / 2) % mod;
  if (y % 2) return (((tmp * tmp) % mod) * x) % mod;
  return (tmp * tmp) % mod;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    f[a[i]]++;
  }
  long long ans = 0;
  for (int j = 0; j <= 20; j++) {
    for (int mask = 0; mask <= 1e6; mask++) {
      if ((mask & (1 << j)) == 0) {
        f[mask] = (f[mask] + f[mask + (1 << j)]) % mod;
        assert(mask + (1 << j) < (1 << 22));
        assert(f[mask] >= 0);
      }
    }
  }
  for (int i = 0; i < (1 << 20); i++) {
    long long mul = 1;
    if (__builtin_popcount(i) % 2 == 1) mul = -1;
    ans = (ans + ((mul * quickPow(2, f[i])) % mod)) % mod;
    ans = (ans + mod * mod) % mod;
    if (ans < 0) {
      break;
    }
  }
  cout << ans;
  return 0;
}