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 long long int mod = 1e9 + 7;
void add(long long int &x, long long int y) { x = (x + y) % mod; }
const int N = 1 << 20;
long long int dp[N];
long long int p2[N];
void init(void) {
  p2[0] = 1;
  for (int i = (int)(1); i < (int)(N); i++) {
    p2[i] = p2[i - 1] * 2 % mod;
  }
}
int main(void) {
  init();
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = (int)(0); i < (int)(n); i++) {
    cin >> a[i];
    dp[a[i]]++;
  }
  for (int i = (int)(0); i < (int)(20); i++) {
    for (int bits = (int)(0); bits < (int)(N); bits++) {
      if (bits & 1 << i) {
        dp[bits ^ 1 << i] += dp[bits];
      }
    }
  }
  for (int bits = (int)(0); bits < (int)(N); bits++) {
    dp[bits] = p2[dp[bits]];
  }
  long long int tot = 0;
  for (int bits = (int)(0); bits < (int)(N); bits++) {
    long long int tmp = dp[bits];
    if (__builtin_popcount(bits) % 2 == 1) {
      tmp = mod - tmp;
    }
    add(tot, tmp);
  }
  cout << tot << endl;
}