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 cnt[1 << 20];
inline long long int pow(long long int x, long long int y, long long int p) {
  long long int ans = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) ans = (ans * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return ans;
}
inline long long int invmod(long long int x, long long int p) {
  return pow(x, p - 2, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, m;
  cin >> n;
  m = 20;
  long long int rev = (1 << m) - 1;
  for (long long int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    cnt[x]++;
  }
  for (long long int i = 0; i < m; i++)
    for (long long int j = rev; j >= 0; j--)
      if ((j & (1 << i)) == 0) cnt[j] += cnt[j ^ (1 << i)];
  long long int ans = 0;
  for (long long int i = 0; i <= rev; i++) {
    long long int x = pow(2, cnt[i], 1000000007) - 1;
    long long int y = (__builtin_popcount(i)) % 2;
    if (y)
      ans = (ans - x) % 1000000007;
    else
      ans = (ans + x) % 1000000007;
  }
  cout << (ans + 1000000007) % 1000000007;
}