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 BITS = 20;
const int MAXN = 1 << BITS;
const long long mod = 1e9 + 7;
void subsets(int* A, int nb) {
  if (nb == 0) return;
  int half = 1 << (nb - 1);
  subsets(A, nb - 1);
  subsets(A + half, nb - 1);
  for (int i = (0); i < int(half); ++i)
    tie(A[i], A[i + half]) = make_tuple(A[i] + A[i + half], A[i + half]);
}
int main(void) {
  static int frek[MAXN];
  memset(frek, 0, sizeof frek);
  int n;
  scanf("%d", &n);
  for (int i = (0); i < int(n); ++i) {
    int x;
    scanf("%d", &x);
    ++frek[x];
  }
  static long long pow2[MAXN];
  for (int i = (0); i < int(MAXN); ++i)
    pow2[i] = (i == 0) ? 1 : pow2[i - 1] * 2 % mod;
  subsets(frek, BITS);
  long long ans = 0;
  for (int st = (0); st < int(MAXN); ++st) {
    int sign = 1;
    for (int i = (0); i < int(BITS); ++i)
      if (st & (1 << i)) sign = -sign;
    ans = (ans + sign * pow2[frek[st]]) % mod;
  }
  printf("%lld\n", (ans % mod + mod) % mod);
  return 0;
}