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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) {
    long long n;
    cin >> n;
    long long currentK = 0;
    long long seen = 0;
    n--;
    long long transformations[3][4] = {
        {0, 1, 2, 3}, {0, 2, 3, 1}, {0, 3, 1, 2}};
    while (true) {
      if (3 * (1ll << currentK) > n) {
        long long solution = 0;
        long long rem = n % 3;
        n /= 3;
        solution += ((rem + 1) << currentK);
        for (int i = 0; i <= currentK + 2; i += 2) {
          solution |= ((transformations[rem % 3][(n >> (i)) & 3]) << i);
        }
        cout << solution << "\n";
        break;
      } else {
        n -= 3 * (1ll << currentK);
        currentK += 2;
      }
    }
  }
  return 0;
}
