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 mod = 1000000007;
int pow(int a, int n) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1LL * res * a % mod;
    a = 1LL * a * a % mod;
    n >>= 1;
  }
  return res;
}
void solve(int f[], int n, int ty) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < (1 << n); j++) {
      if (j & (1 << i)) continue;
      if ((f[j | (1 << i)] += ty * f[j]) >= mod) f[j | (1 << i)] -= mod;
      if ((f[j | (1 << i)]) < 0) f[j | (1 << i)] += mod;
    }
  }
}
int f[1 << 20];
int main() {
  int n;
  scanf("%d", &n);
  int mask = (1 << 20) - 1;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    f[x ^ mask]++;
  }
  solve(f, 20, 1);
  for (int i = 0; i < (1 << 20); i++) {
    f[i] = pow(2, f[i]);
  }
  solve(f, 20, -1);
  printf("%d\n", f[mask]);
  return 0;
}