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 = 2e6 + 10;
const int B = 20;
long long mod = 1e9 + 7;
long long po[N];
long long f[N];
int main() {
  int n;
  scanf("%d", &n);
  po[0] = 1;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    f[x]++;
    po[i] = (po[i - 1] << 1) % mod;
  }
  for (int i = 0; i < B; i++)
    for (int mask = 0; mask < (1 << B); mask++)
      if (!(mask & (1 << i))) f[mask] += f[mask ^ (1 << i)];
  long long ans = 0;
  for (int mask = 1; mask < (1 << B); mask++) {
    long long cur = po[f[mask]] - 1 + mod;
    cur %= mod;
    int cnt = __builtin_popcount(mask);
    if (!(cnt & 1)) cur = mod - cur;
    ans = (ans + cur) % mod;
  }
  ans = po[n] - 1 + mod - ans;
  ans %= mod;
  printf("%I64d\n", ans);
}