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 N = (int)1e6 + 9, K = 20, M = 1 << K, mod = (int)1e9 + 7;
int n, a[N], f[M], pw[N] = {1};
short builtin_popcount(int x) {
  short ans = 0;
  while (x) {
    ans += (x & 1);
    x >>= 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1, a; i <= n; ++i) {
    cin >> a;
    ++f[a];
    pw[i] = pw[i - 1] << 1;
    if (pw[i] > mod) pw[i] -= mod;
  }
  for (int i = 0; i < K; ++i)
    for (int mask = 0; mask < M; ++mask)
      if (!(mask & (1 << i))) f[mask] += f[mask ^ (1 << i)];
  long long ans = 0;
  for (int mask = 0; mask < M; ++mask) {
    if (builtin_popcount(mask) & 1)
      ans -= pw[f[mask]];
    else
      ans += pw[f[mask]];
  }
  ans %= mod;
  if (ans < 0) ans += mod;
  cout << ans;
}