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;
int n, t[1111111], u[1111111];
int p2[1111111];
long long ans = 0;
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) {
    int a;
    scanf("%d", &a);
    t[a]++;
  }
  for (int i = (0); i <= (19); i++) {
    for (int j = (0); j <= (((1 << 20) - 1)); j++) {
      u[j] += t[j];
      if (j & (1 << i)) u[j - (1 << i)] += t[j];
    }
    for (int j = (0); j <= (((1 << 20) - 1)); j++) {
      t[j] = u[j];
      u[j] = 0;
    }
  }
  p2[0] = 1;
  for (int i = (1); i <= (n); i++) p2[i] = p2[i - 1] * 2 % 1000000007;
  int res = 0;
  for (int i = (0); i <= (((1 << 20) - 1)); i++) {
    int cb = 0;
    for (int j = (0); j <= (19); j++) cb += (i & (1 << j)) >> j;
    if (cb % 2 == 0)
      res += p2[t[i]] - 1;
    else
      res -= p2[t[i]] - 1;
    res = (res % 1000000007 + 1000000007) % 1000000007;
  }
  printf("%d\n", res);
  return 0;
}