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 = 22;
const int mod = 1e9 + 7;
int dp[1 << N];
int pw[1 << N];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int cur;
    cin >> cur;
    dp[cur]++;
  }
  pw[0] = 1;
  for (int i = 1; i < (1 << N); i++) pw[i] = (long long)pw[i - 1] * 2 % mod;
  for (int j = 0; j < N; j++)
    for (int i = 0; i < (1 << N); i++)
      if (i & (1 << j)) dp[i ^ (1 << j)] += dp[i];
  int ans = 0;
  for (int i = 0; i < (1 << N); i++) {
    if (__builtin_popcount(i) & 1)
      ans = (ans + mod - pw[dp[i]]) % mod;
    else
      ans = (ans + pw[dp[i]]) % mod;
  }
  cout << ans << endl;
  return 0;
}