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 N = 2e6 + 5, hgs = 1e9 + 7, K = 19, S = (1 << 20) - 1;
int n, x, f[N], un[N];
void Solution(int Base) {
  for (int k = 0; k <= K; ++k)
    for (int i = 0; i <= S; ++i)
      if (!(i & (1 << k))) f[i] = (f[i] + f[i | (1 << k)] * Base + hgs) % hgs;
}
signed main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &x), ++f[x];
  for (int i = 1; i <= n; ++i) un[i] = (un[i - 1] << 1 | 1) % hgs;
  Solution(1);
  for (int i = 0; i <= S; ++i) f[i] = un[f[i]];
  Solution(-1);
  printf("%d\n", f[0]);
  return 0;
}