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;
using Int = long long;
int PREP = (cin.tie(nullptr), ios::sync_with_stdio(false),
            cout << fixed << setprecision(9), 0);
Int fast(Int N) {
  Int ans = 0, t;
  for (t = 0; t < 32; t++)
    if (N <= (1LL << (2 + 2 * t)) - 1) break;
  const int A[3] = {3, 1, 2};
  N -= (1LL << (2 * t));
  Int a = N % 3;
  ans += (a + 1) << (2 * t);
  Int d = 1;
  N /= 3;
  for (int i = 0; i < t; i++) {
    Int b = N % 4;
    if (b != 0) {
      ans += A[(a + b) % 3] * d;
    }
    N /= 4;
    d *= 4;
  }
  return ans;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    Int N;
    cin >> N;
    cout << fast(N) << '\n';
  }
  return 0;
}
