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 Mod = 1000000007;
int f[1 << 20];
int p[1 << 20];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    f[x] += 1;
  }
  for (int i = 1; i < (1 << 20); i *= 2) {
    for (int j = (1 << 20) - 1; j >= 0; --j) {
      if ((j & i) != 0) {
        f[j ^ i] += f[j];
      }
    }
  }
  int ans = 0;
  p[0] = 1;
  for (int i = 1; i < (1 << 20); ++i) p[i] = (p[i - 1] * 2) % Mod;
  for (int i = 0; i < (1 << 20); ++i) {
    int cnt = __builtin_popcount(i);
    if (cnt & 1)
      ans -= p[f[i]];
    else
      ans += p[f[i]];
    ans %= Mod;
  }
  ans += Mod, ans %= Mod;
  printf("%d\n", ans);
  return 0;
}