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 cnt[(1 << 20)];
int pwr[1000005];
int mod_mul(long long a, long long b) { return (a * b) % 1000000007; }
int mod_plus(int a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  return a;
}
int mod_min(int a, int b) {
  a -= b;
  if (a < 0) a += 1000000007;
  return a;
}
void go(int st, int total) {
  if (!total) return;
  for (int i = 0; i < total; i++) cnt[st + i] += cnt[st + i + total];
  go(st, total / 2);
  go(st + total, total / 2);
}
int main() {
  int i, n, j;
  pwr[0] = 1;
  for (i = 1; i < 1000005; i++) pwr[i] = mod_mul(pwr[i - 1], 2);
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &j);
    cnt[j]++;
  }
  go(0, (1 << 20) / 2);
  int ans = 0;
  for (i = 0; i < (1 << 20); i++) {
    if (__builtin_popcount(i) % 2)
      ans = mod_min(ans, pwr[cnt[i]]);
    else
      ans = mod_plus(ans, pwr[cnt[i]]);
  }
  cout << ans << endl;
  return 0;
}