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 mod = 1e9 + 7;
const int N = 1e6;
int cnt[N + 10];
long long two[N + 10];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    cnt[a]++;
  }
  for (int j = 19; j >= 0; j--) {
    for (int i = 1; i <= N; i++) {
      if ((i >> j) & 1) {
        cnt[i - (1 << j)] += cnt[i];
      }
    }
  }
  two[0] = 1;
  for (int i = 1; i <= N; i++) {
    two[i] = (two[i - 1] * 2) % mod;
  }
  for (int i = 0; i <= N; i++) {
    two[i] = (two[i] - 1 + mod) % mod;
  }
  long long ans = two[n];
  for (int i = 1; i <= N; i++) {
    int one = __builtin_popcount(i);
    if (one & 1) {
      ans = (ans - two[cnt[i]] + mod) % mod;
    } else {
      ans = (ans + two[cnt[i]]) % mod;
    }
  }
  printf("%lld\n", ans);
}