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 maxn = 1 << 20;
const int mod = 1e9 + 7;
int A[maxn], dp[maxn];
int main(void) {
  int n, a;
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &a), dp[a]++;
  int all = (1 << 20);
  for (int i = 0; i < 20; i++) {
    for (int s = all - 1; s >= 0; s--) {
      if ((s >> i & 1) == 0) dp[s] += dp[s | (1 << i)];
    }
  }
  A[0] = 1;
  for (int i = 1; i < maxn; i++) A[i] = A[i - 1] * 2 % mod;
  long long res = 0;
  for (int i = 0; i < all; i++) {
    int t = 1;
    for (int j = 0; j < 20; j++) {
      if (i & (1 << j)) t = -t;
    }
    res = (res + t * (A[dp[i]] - 1) + mod) % mod;
  }
  cout << res;
}