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 MV = 1048576, ML = 20, MOD = 1e9 + 7;
int N, freq[MV], sup[MV];
long long int pow2(int n) {
  long long int base = 2, res = 1;
  while (n) {
    if (n & 1) res = res * base % MOD;
    base = base * base % MOD;
    n /= 2;
  }
  return res;
}
int main() {
  int n;
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &n);
    ++freq[n];
  }
  for (int m = 0; m < MV; m++) sup[m] = freq[m];
  for (int i = 0; i < 20; i++)
    for (int m = 0; m < MV; m++)
      if (!(m & (1 << i))) sup[m] += sup[m ^ (1 << i)];
  long long int ans = 0;
  for (int m = 1; m < MV; m++) {
    if (__builtin_popcount(m) & 1)
      ans = (ans + pow2(sup[m]) - 1) % MOD;
    else
      ans = (ans + 1 + MOD - pow2(sup[m])) % MOD;
  }
  printf("%lld\n", (pow2(N) + MOD - ans - 1) % MOD);
}