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;
int N;
int A[1 << 20], f[1 << 20], g[1 << 20];
long long ans, pw[1 << 20];
int main(void) {
  scanf("%d", &N);
  pw[0] = 1;
  for (int i = 1; i <= N; i++) {
    pw[i] = (pw[i - 1] * 2) % mod;
  }
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
    f[A[i]]++;
  }
  for (int i = 0; i < 20; i++) {
    for (int j = (1 << 20) - 1; j >= 0; j--) {
      if ((j & (1 << i)) == 0) {
        f[j] += f[j | (1 << i)];
      }
    }
  }
  for (int i = 0; i < (1 << 20); i++) {
    g[i] = __builtin_popcount(i);
    if (g[i] & 1)
      g[i] = -1;
    else
      g[i] = 1;
    ans += g[i] * pw[f[i]];
    ans %= mod;
    if (ans < 0) ans += mod;
  }
  printf("%lld\n", ans);
}