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 dp[(1 << 21)][21];
int n;
int arr[1000001];
int power(int x, int y) {
  if (y == 0) return 1;
  int a = power(x, y / 2);
  a = (1LL * a * a) % mod;
  if (y % 2) a = (1LL * a * x) % mod;
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  for (int i = 1; i <= n; i++) dp[arr[i]][0]++;
  int tot = 0;
  for (int k = 1; k <= 20; k++) {
    for (int i = 0; i < (1 << 20); i++) {
      dp[i][k] = dp[i][k - 1];
      if ((i & (1 << (k - 1))) == 0) dp[i][k] += dp[i + (1 << (k - 1))][k - 1];
    }
  }
  for (int i = 0; i <= 1000000; i++) {
    if (dp[i][20] == 0) continue;
    int val = power(2, dp[i][20]);
    val--;
    if (val < 0) val += mod;
    if (__builtin_popcount(i) % 2) {
      tot -= val;
      if (tot < 0) tot += mod;
    } else {
      tot += val;
      tot %= mod;
    }
  }
  cout << tot;
}