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 nmax = (1 << 20) + 5;
const int mod = 1e9 + 7;
int n, i, j, x, lim, sol;
int f[nmax], cnt[nmax], two[nmax];
void fix(int &x) {
  if (x < 0) x += mod;
  if (x >= mod) x -= mod;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    f[x]++;
  }
  lim = 1 << 20;
  two[0] = 1;
  for (i = 1; i < lim; i++) two[i] = (two[i - 1] * 2) % mod;
  for (i = 0; i < 20; i++)
    for (j = 0; j < lim; j++) {
      if ((1 << i) & j)
        cnt[j]++;
      else
        f[j] += f[j + (1 << i)];
    }
  for (i = 1; i < lim; i++) {
    if (cnt[i] & 1) {
      sol += two[f[i]] - 1;
      fix(sol);
    } else {
      sol -= two[f[i]] - 1;
      fix(sol);
    }
  }
  sol = two[n] - 1 - sol;
  fix(sol);
  printf("%d\n", sol);
  return 0;
}