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 inf = (int)1e9;
const double pi = acos(-1.0);
const double eps = 1e-9;
const int mod = inf + 7;
int n, x, w[2000100], p[2000100];
long long res;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    w[x] = w[x] + 1;
  }
  for (int i = 0; i <= 20; i++)
    for (int mask = 0; mask < (1 << 20); mask++)
      if (mask & (1 << i)) w[mask ^ (1 << i)] += w[mask];
  p[0] = 1;
  for (int i = 1; i <= n; i++) p[i] = p[i - 1] * 2 % mod;
  for (int mask = 0; mask < (1 << 20); mask++) {
    int c = 1;
    for (int i = 0; i < 20; i++)
      if (mask & (1 << i)) c = -c;
    res = (res + c * p[w[mask]] + mod) % mod;
  }
  printf("%I64d", res);
  return 0;
}