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 = 1e9 + 7;
const int maxn = 1 << 20;
int n, a[maxn], f[21][maxn], p2[maxn], ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    f[0][a[i]]++;
  }
  for (int i = 1; i <= 20; i++) {
    for (int j = 0; j < (1 << 20); j++) {
      if ((j >> (i - 1)) & 1) {
        f[i][j] = f[i - 1][j];
      } else {
        f[i][j] = f[i - 1][j] + f[i - 1][j + (1 << (i - 1))];
        f[i][j] -= (f[i][j] >= mod) ? mod : 0;
      }
    }
  }
  p2[0] = 1;
  for (int i = 1; i <= n; i++)
    p2[i] = p2[i - 1] + p2[i - 1], p2[i] -= (p2[i] >= mod) ? mod : 0;
  for (int i = 0; i < (1 << 20); i++)
    if (__builtin_popcount(i) & 1) {
      ans -= p2[f[20][i]];
      if (ans < 0) ans += mod;
    } else {
      ans += p2[f[20][i]];
      ans -= (ans >= mod) ? mod : 0;
    }
  printf("%d", ans);
}