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;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  while (y) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y / 2, x = (x * x) % 1000000007;
  }
  return res % 1000000007;
}
long long int dp[1 << 22];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  long long int i, j;
  for (i = 0; i < n; i++) {
    long long int k;
    cin >> k;
    dp[k]++;
  }
  for (i = 0; i < 20; i++) {
    for (j = 0; j < (1 << 20); j++) {
      if (!(j & (1 << i))) dp[j] += (dp[j ^ (1 << i)]);
    }
  }
  for (i = 0; i < (1 << 20); i++) {
    dp[i] = (power(2, dp[i]) - 1 + 1000000007) % 1000000007;
  }
  for (i = 0; i < 20; i++) {
    for (j = 0; j < (1 << 20); j++) {
      if (!(j & (1 << i)))
        dp[j] = (dp[j] - (dp[j ^ (1 << i)]) + 1000000007) % 1000000007;
    }
  }
  cout << dp[0];
}