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 int mod = 1000000007;
int dp[(1 << 20) - 1];
int g[(1 << 20) - 1];
long long int modPow(int a, int n) {
  if (n == 0) return 1;
  if (n % 2 == 0)
    return (modPow(a, n / 2) * modPow(a, n / 2)) % mod;
  else
    return (modPow(a, n / 2) * modPow(a, n / 2) * a) % mod;
  ;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    dp[x]++;
  }
  for (int i = 0; i < 20; i++) {
    for (int mask = (1 << 20) - 2; mask >= 0; mask--) {
      if (mask & (1 << i)) {
        dp[mask ^ (1 << i)] += dp[mask];
      }
    }
  }
  for (int i = 0; i < (1 << 20) - 1; i++) {
    g[i] = __builtin_popcount(i);
  }
  long long int ans = 0;
  for (int i = 0; i < (1 << 20) - 1; i++) {
    ans += (long long int)pow(-1, g[i]) * (modPow(2, dp[i]) - 1);
    ans = (ans + mod) % mod;
  }
  cout << ans << "\n";
}