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 dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
int kx8[] = {1, 1, 2, 2, -1, -1, -2, -2};
int ky8[] = {2, -2, 1, -1, 2, -2, 1, -1};
const int M = (1 << 20) + 5;
long long dp[M];
long long relax(long long val) {
  if (val >= 1000000007) val -= 1000000007;
  if (val >= 1000000007) val %= 1000000007;
  if (val < 0) val += 1000000007;
  return val;
}
long long pwr[M];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, N, m;
  cin >> m;
  for (int i = 0; i < (m); i++) {
    int a;
    cin >> a;
    dp[a]++;
  }
  pwr[0] = 1;
  for (int i = 1; i <= m; i++) {
    pwr[i] = (pwr[i - 1] * 2LL) % 1000000007;
  }
  N = 20;
  n = (1 << N);
  for (int i = 0; i < N; ++i) {
    for (int mask = n - 1; mask >= 0; --mask) {
      if (!(bool)(mask & (1 << i))) {
        dp[mask] += dp[mask ^ (1 << i)];
      }
    }
  }
  long long res = 0;
  for (long long i = 0; i < n; i++) {
    if (__builtin_popcountll(i) % 2) {
      res -= pwr[dp[i]];
      res = relax(res);
    } else {
      res += pwr[dp[i]];
      res = relax(res);
    }
  }
  cout << res << endl;
  return 0;
}