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 = (1 << 20) + 5;
const int MO = 1e9 + 7;
int a[N], cnt[N], pw[N] = {1};
void add(int &x, int y) {
  x += y;
  if (x >= MO) x -= MO;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = int(1); i <= int(n); i++) pw[i] = pw[i - 1], add(pw[i], pw[i]);
  for (int i = int(1); i <= int(n); i++) scanf("%d", a + i), cnt[a[i]]++;
  for (int i = int(0); i <= int((20) - 1); i++) {
    for (int j = int(0); j <= int((1 << 20) - 1); j++) {
      if ((j & (1 << i)) == 0) {
        cnt[j] += cnt[j + (1 << i)];
      }
    }
  }
  int ans = 0;
  for (int i = int(0); i <= int((1 << 20) - 1); i++) {
    if (__builtin_parity(i)) {
      add(ans, MO - pw[cnt[i]]);
    } else {
      add(ans, pw[cnt[i]]);
    }
  }
  cout << ans << endl;
  return 0;
}