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 + 5;
const int mod = 1e9 + 7;
const int N = 20;
long long a[maxn], dp[maxn], p[maxn];
int main() {
  p[0] = 1;
  for (int i = 1; i < maxn; i++) {
    p[i] = p[i - 1] * 2 % mod;
  }
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    dp[a[i]]++;
  }
  int s = 1 << N;
  for (int j = 0; j < N; j++)
    for (int i = 0; i < s; i++)
      if (!(i >> j & 1)) dp[i] += dp[i ^ (1 << j)];
  for (int i = 0; i < s; i++) {
    dp[i] = (p[dp[i]] - 1 + mod) % mod;
  }
  long long ans = 0;
  for (int i = 0; i < s; i++) {
    int cnt = 0;
    for (int j = 0; j < N; j++) {
      cnt += (i >> j & 1);
    }
    long long f = (cnt % 2 ? -1 : 1);
    ans = (ans + f * dp[i] % mod + mod) % mod;
  }
  cout << ans << endl;
  return 0;
}