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;
using ll = long long;
using pii = pair<int, int>;
const int INF = 0x3f3f3f3f, N = 1e6 + 5, bit = 20, mod = 1e9 + 7;
int n, a[N];
ll dp[N << 1];
ll powmod(ll a, ll b, ll mod) {
  ll ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    dp[a[i]]++;
  }
  for (int i = 0; i < bit; i++) {
    for (int mask = 0; mask < (1 << bit); mask++) {
      if (mask & (1 << i))
        dp[mask ^ (1 << i)] = (dp[mask ^ (1 << i)] + dp[mask]) % mod;
    }
  }
  ll ans = 0;
  for (int i = 0; i < (1 << bit); i++) {
    int cnt = __builtin_popcount(i);
    if (cnt & 1)
      ans = (ans - powmod(2, dp[i], mod) + 1 + mod) % mod;
    else
      ans = (ans + powmod(2, dp[i], mod) - 1 + mod) % mod;
  }
  cout << (ans + mod) % mod << '\n';
  return 0;
}