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 sz = 20, mod = 1e9 + 7;
int dp[1 << sz], st[1 << sz];
int main() {
  int n, an = 0;
  cin >> n;
  for (int a = 0; a < n; a++) {
    int x;
    scanf("%d", &x);
    dp[x]++;
    st[a + 1] = (st[a] * 2 + 1) % mod;
  }
  for (int i = 0; i < sz; i++) {
    for (int mask = 0; mask < (1 << sz); mask++) {
      if ((mask & (1 << i)) == 0) dp[mask] += dp[mask + (1 << i)];
    }
  }
  for (int mask = 0; mask < (1 << sz); mask++) {
    int q = __builtin_popcount(mask);
    if (q & 1)
      an -= st[dp[mask]];
    else
      an += st[dp[mask]];
    if (an < 0) an += mod;
    if (an >= mod) an -= mod;
  }
  cout << an;
}