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 N;
long long a[1000000];
long long dp[1 << 22];
const int MOD = 1e9 + 7;
int main() {
  cin.sync_with_stdio(false);
  while (cin >> N) {
    for (int i = 0; i < N; ++i) cin >> a[i];
    fill(dp, dp + (1 << 20), 1);
    for (int i = 0; i < N; ++i) dp[a[i]] = (dp[a[i]] * 2) % MOD;
    for (int i = 0; i < 20; ++i) {
      for (int j = 0; j < (1 << 20); ++j) {
        if ((j >> i) & 1) continue;
        dp[j] = (dp[j] * dp[j | (1 << i)]) % MOD;
      }
    }
    long long ans = 0;
    for (int i = 0; i < (1 << 20); ++i) {
      int cnt = __builtin_popcount(i);
      if (cnt % 2 == 0)
        ans = (ans + dp[i]) % MOD;
      else
        ans = (ans - dp[i] + MOD) % MOD;
    }
    cout << ans << endl;
  }
  return 0;
}