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 maxn = 2e6 + 7;
const int inf = 2e9;
const long long inff = 1e18;
const long long mod = 1e9 + 7;
int pow2[maxn];
int dp[maxn];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  pow2[0] = 1;
  for (int i = 1; i <= 1000000; i++) pow2[i] = pow2[i - 1] * 2 % mod;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    dp[x]++;
  }
  for (int j = 0; j < 20; j++)
    for (int i = 0; i < (1 << 20); i++) {
      if ((i >> j) & 1 ^ 1) dp[i] += dp[i ^ (1 << j)];
    }
  int ans = 0;
  for (int i = 0; i < (1 << 20); i++) {
    if (__builtin_popcount(i) & 1)
      ans = (ans + mod - pow2[dp[i]]) % mod;
    else
      ans = (ans + pow2[dp[i]]) % mod;
  }
  cout << ans << '\n';
}