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 maxn = (1 << 20);
int cnt[maxn];
long long f[maxn];
const long long MOD = 1e9 + 7;
long long pw(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % MOD;
    b >>= 1;
    a = a * a % MOD;
  }
  return res;
}
int sign(int i) {
  if (i & 1) return -1;
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int a;
    cin >> a;
    ++cnt[a];
  }
  for (int i = 0; i <= 1000000; ++i) f[i] = cnt[i];
  for (int b = 0; b < 20; ++b) {
    for (int i = 0; i <= 1000000; ++i) {
      if (!(i & (1 << b))) f[i] += f[i ^ (1 << b)];
    }
  }
  long long ans = 0;
  for (int i = 0; i <= 1000000; ++i) {
    ans += sign(__builtin_popcount(i)) * (pw(2, f[i]) - 1);
    if (ans < 0) ans += MOD;
    if (ans >= MOD) ans -= MOD;
  }
  cout << ans;
}