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>
int a[1000100];
const int mod = 1000000007;
int b[1 << 20];
int p[1000100];
int main() {
  int n;
  scanf("%d", &n);
  int C = (1 << 20) - 1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[a[i]]++;
  }
  for (int i = 0; i < 20; i++) {
    for (int j = 0; j < 1 << 20; j++) {
      if ((j >> i & 1) == 0) {
        b[j] += b[j | (1 << i)];
      }
    }
  }
  int ans = 0;
  p[0] = 1;
  for (int i = 0; i < n; i++) {
    p[i + 1] = (p[i] << 1) % mod;
  }
  for (int i = 0; i < 1 << 20; i++) {
    int t = p[b[i]];
    for (int j = 0; j < 20; j++) {
      if (i >> j & 1) t = -t;
    }
    if (t < 0) {
      t += mod;
    }
    ans += t;
    if (ans >= mod) {
      ans -= mod;
    }
  }
  printf("%d\n", ans);
  return 0;
}