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 = (1 << 20) + 100;
bool one(int mask, int i) { return mask & (1 << i); }
int mod = 1000000007;
int n, a[N], d[N];
int p[N];
int cnt[N];
int was[N];
int main() {
  scanf("%d", &n);
  p[0] = 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    p[i] = p[i - 1] << 1;
    if (p[i] >= mod) p[i] -= mod;
    d[a[i]]++;
  }
  for (int i = 0; i <= n; ++i) p[i] = (p[i] - 1) % mod;
  for (int i = 20; i >= 0; --i) {
    for (int j = (1 << i); j < (1 << 20); ++j)
      if (one(j, i)) d[j - (1 << i)] += d[j];
  }
  long long ans = 0;
  for (int i = 1; i <= 1000000; ++i) {
    if (!d[i]) continue;
    int kol = __builtin_popcount(i);
    if (kol & 1)
      ans = (ans + p[d[i]]) % mod;
    else
      ans = (ans - p[d[i]] + mod) % mod;
  }
  ans = (p[n] - ans) % mod;
  ans = (ans + mod) % mod;
  cout << ans;
  return 0;
}