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 MAXLA = 20, MOD = 1e9 + 7;
int n, x, f[(1 << MAXLA)], ans;
int powmod(int b, int e) {
  if (!e) return 1;
  if (e & 1) return (1LL * b * powmod(b, e - 1)) % MOD;
  int aux = powmod(b, e >> 1);
  return (1LL * aux * aux) % MOD;
}
int main() {
  for (scanf("%d", &n); n--;) scanf("%d", &x), f[x]++;
  for (int i = 0, mask; i < MAXLA; ++i)
    for (mask = 0; mask < (1 << MAXLA); ++mask)
      if (mask & (1 << i)) f[mask ^ (1 << i)] += f[mask];
  for (int mask = 0; mask < (1 << MAXLA); ++mask)
    ans =
        (0LL + ans +
         1LL * powmod(MOD - 1, __builtin_popcount(mask)) * powmod(2, f[mask])) %
        MOD;
  printf("%d\n", ans);
}