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 N = 1005000;
const int BASE = 1000000007;
int n, pw[N], f[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    f[x]++;
  }
  pw[0] = 1;
  for (int i = 1; i < N; i++) {
    pw[i] = pw[i - 1] + pw[i - 1];
    if (pw[i] >= BASE) {
      pw[i] -= BASE;
    }
  }
  for (int j = 0; j < 20; j++) {
    for (int i = N - 1; i >= 0; i--) {
      if (((i & (1 << j)) == 0) && (i + (1 << j) < N)) {
        f[i] += f[i + (1 << j)];
      }
    }
  }
  for (int i = 0; i < N; i++) {
    f[i] = pw[f[i]];
  }
  int ans = 0;
  for (int i = 0; i < N; i++) {
    int o = 0;
    int x = i;
    while (x > 0) {
      o ^= 1;
      x &= (x - 1);
    }
    if (o) {
      ans -= f[i];
      if (ans < 0) {
        ans += BASE;
      }
    } else {
      ans += f[i];
      if (ans >= BASE) {
        ans -= BASE;
      }
    }
  }
  printf("%d\n", ans);
}