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], pow2[1000009];
void pre() {
  pow2[0] = 1;
  for (int i = int(1); i <= int(1000000); i++)
    pow2[i] = (pow2[i - 1] * 2) % 1000000007;
}
int main() {
  pre();
  int n;
  scanf("%d", &n);
  for (int i = int(0); i <= int(n - 1); i++) {
    int p;
    scanf("%d", &p);
    cnt[p]++;
  }
  for (int i = int(0); i <= int(19); i++) {
    for (int j = int((1 << 20) - 1); j >= int(0); j--) {
      if (j & (1 << i)) cnt[j - (1 << i)] += cnt[j];
    }
  }
  int ans = 0;
  for (int i = int(0); i <= int((1 << 20) - 1); i++) {
    int bits = 0, temp = i;
    while (temp > 0) {
      bits++;
      temp -= temp & -temp;
    }
    if (bits % 2 == 0)
      ans = (ans + pow2[cnt[i]]) % 1000000007;
    else
      ans = (ans - pow2[cnt[i]] + 1000000007) % 1000000007;
  }
  printf("%d\n", ans);
  return 0;
}