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 two[(1 << 20)];
int A[(1 << 20)];
int S[(1 << 20)];
int main() {
  two[0] = 1;
  for (int i = (1); i <= ((1 << 20) - 1); ++i)
    two[i] = two[i - 1] * 2 % 1000000007;
  int N;
  scanf("%d", &N);
  for (int i = (0); i < (N); ++i) scanf("%d", A + i), S[A[i]]++;
  for (int b = (0); b < (20); ++b)
    for (int s = (0); s < ((1 << 20)); ++s)
      if (s & (1 << b)) S[s - (1 << b)] += S[s];
  int ans = 0;
  for (int s = (0); s < ((1 << 20)); ++s)
    if (__builtin_popcount(s) & 1)
      ans = (ans - two[S[s]] + 1000000007) % 1000000007;
    else
      ans = (ans + two[S[s]]) % 1000000007;
  cout << ans << endl;
}