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 a[2000009];
long long int fac[2000009];
long long int power(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    res %= 1000000007;
    y = y >> 1;
    x = x * x;
    x %= 1000000007;
  }
  return res;
}
long long int inv(long long int x) { return power(x, 1000000007 - 2); }
long long int ncr(long long int n, long long int r) {
  long long int res = fac[n];
  res *= inv(fac[r]);
  res %= 1000000007;
  res *= inv(fac[n - r]);
  res %= 1000000007;
  return res;
}
void solve() {
  long long int n;
  cin >> n;
  fac[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i;
    fac[i] %= 1000000007;
  }
  long long int p = power(2, 19);
  long long int x;
  for (long long int i = 0; i < n; i++) {
    cin >> x;
    a[x]++;
  }
  long long int ans = power(2, n);
  ans--;
  long long int z = power(2, 20);
  z--;
  for (long long int j = 0; j < 20; j++) {
    long long int b = 0;
    for (long long int i = 1; i <= z; i++) {
      long long int x = i | ((long long int)1 << j);
      if (x != i) a[i] += a[x];
    }
  }
  for (long long int i = z; i > 0; i--) {
    long long int b = 0;
    for (long long int j = 0; j < 20; j++) {
      long long int x = i | ((long long int)1 << j);
      if (x == i) b++;
    }
    long long int t = power(2, a[i]);
    t--;
    t %= 1000000007;
    if (b % 2 == 0) {
      ans += t;
    } else {
      ans -= t;
      ans %= 1000000007;
      ans += 1000000007;
    }
    ans %= 1000000007;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  for (long long int i = 0; i < t; i++) {
    solve();
  }
}