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 N = 2e6 + 7, mod = 1e9 + 7, Lg = 20;
inline int add(int a, int b) {
  int c = a + b;
  if (c >= mod) c -= mod;
  return c;
}
int t[N], a[N];
int tav[N];
int main() {
  tav[0] = 1;
  for (int i = 1; i < N; i++) tav[i] = add(tav[i - 1], tav[i - 1]);
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    t[a[i]]++;
  }
  for (int i = Lg - 1; i >= 0; i--)
    for (int mask = 0; mask < (1 << Lg); mask++)
      if (((mask >> i) & 1) == 0) t[mask] = add(t[mask], t[mask ^ (1 << i)]);
  int ans = 0;
  for (int i = 0; i < (1 << Lg); i++)
    if (__builtin_popcount(i) & 1)
      ans = add(ans, -tav[t[i]] + mod);
    else
      ans = add(ans, tav[t[i]]);
  cout << ans << endl;
  return 0;
}