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;
int n, f[1 << 20];
int POW[1000009];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    f[x]++;
  }
  for (int i = 1; i <= 20; i++)
    for (int j = 0; j < 1 << 20; j++) {
      if ((j & (1 << (i - 1))) == 0) f[j] += f[j + (1 << (i - 1))];
    }
  POW[0] = 1;
  for (int i = 1; i <= n; i++) POW[i] = (long long)POW[i - 1] * 2 % 1000000007;
  long long ans = POW[n];
  for (int i = 1; i < (1 << 20); i++) {
    int cnt = 0;
    for (int j = 0; j < 20; j++) {
      if (i & (1 << j)) cnt++;
    }
    if (cnt % 2 == 1)
      ans -= POW[f[i]];
    else
      ans += POW[f[i]];
    ans %= 1000000007;
    ans += 1000000007;
    ans %= 1000000007;
  }
  cout << ans << endl;
}