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>
#pragma GCC optimize("O2")
using namespace std;
const int MAXN = 1e6 + 10;
const long long MOD = 1e9 + 7;
const long long MOD2 = 998244353;
const long long INF = 8e18;
const int LOG = 22;
long long pw(long long a, long long b, long long mod) {
  return (!b ? 1
             : (b & 1 ? (a * pw(a * a % mod, b / 2, mod)) % mod
                      : pw(a * a % mod, b / 2, mod)));
}
int v[MAXN], dp[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    dp[v[i]]++;
  }
  for (int i = 0; i < LOG; i++) {
    for (int mask = MAXN - 1; ~mask; mask--) {
      if (mask & (1 << i)) {
        dp[mask ^ (1 << i)] += dp[mask];
      }
    }
  }
  long long ans = 0;
  for (int mask = 0; mask < MAXN; mask++) {
    int f = __builtin_popcount(mask);
    int z = ((f % 2) ? -1 : 1);
    ans = (ans + z * pw(2, dp[mask], MOD) + MOD) % MOD;
  }
  cout << ans;
  return 0;
}