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 mo = 1000000007;
const int N = (1 << 20) + 5;
int f[N], g[N];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  g[0] = 1;
  for (int i = 1; i <= (1 << 20); ++i) g[i] = g[i - 1] * 2 % mo;
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    ++f[x];
  }
  for (int j = 1; j < (1 << 20); j <<= 1)
    for (int i = 0; i < (1 << 20); ++i)
      if (i & j) f[i ^ j] += f[i];
  for (int i = 1; i < (1 << 20); ++i) f[i] = g[f[i]];
  int ans = 0;
  for (int j = 1; j < (1 << 20); j <<= 1)
    for (int i = 0; i < (1 << 20); ++i)
      if (i & j) (f[i ^ j] -= f[i]) %= mo;
  for (int i = 1; i < (1 << 20); ++i) (ans += f[i]) %= mo;
  (ans = g[n] - ans) %= mo;
  if (ans < 0) ans += mo;
  cout << ans << endl;
}