Consider the infinite sequence s of positive integers, created by repeating the following steps:

  1. Find the lexicographically smallest triple of positive integers (a, b, c) such that 
    * a ⊕ b ⊕ c = 0, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). 
    * a, b, c are not in s. 
Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2]. 
  2. Append a, b, c to s in this order. 
  3. Go back to the first step. 



You have integer n. Find the n-th element of s.

You have to answer t independent test cases.

A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.

Each of the next t lines contains a single integer n (1≤ n ≤ 10^{16}) — the position of the element you want to know.

Output

In each of the t lines, output the answer to the corresponding test case.

Example

Input


9
1
2
3
4
5
6
7
8
9


Output


1
2
3
4
8
12
5
10
15

Note

The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ... 
#include <bits/stdc++.h>
using namespace std;
long long int T, n, res, ex, res2, con;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> T;
  for (int t = 0; t < T; t++) {
    cin >> n;
    n--;
    res = n % 3;
    ex = res;
    res++;
    n /= 3;
    res2 = 0;
    con = 0;
    while (n != 0) {
      n--;
      res = (res << 2);
      if (n % 4 != 0) {
        res2 += ((((n % 4) + ex - 1) % 3) + 1) << 2 * con;
      }
      n /= 4;
      con++;
    }
    cout << res + res2 << "\n";
  }
}
