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;
long long n, m, x, ans, i, j, md, f[1000002], d[1000002];
int main() {
  cin >> n;
  md = 1000000007;
  m = 1000000;
  f[0] = 1;
  for (i = 1; i <= m; i++) f[i] = (f[i - 1] * 2) % md;
  for (i = 1; i <= n; i++) {
    cin >> x;
    d[x]++;
  }
  for (i = 20; i >= 0; i--)
    for (j = m; j >= 0; j--)
      if (j & (1 << (i))) d[j ^ (1 << i)] += d[j];
  for (int j = 0; j <= m; j++)
    if (d[j]) {
      if (__builtin_popcount(j) % 2)
        ans = (ans - f[d[j]] + 1 + md) % md;
      else
        ans = (ans + f[d[j]] - 1 + md) % md;
    }
  cout << (ans + md) % md << "\n";
}