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;
const int N = 2e5 + 2;
const int mod = 1e9 + 7;
long long v[] = {0, 2, 3, 1};
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long x = ceil(n / 3.0);
    long long go = 1, cur = 1;
    while (cur < x) {
      go *= 4;
      cur += go;
    }
    cur -= go;
    long long limInf = go;
    long long limSup = 2 * go - 1;
    long long A = limInf + (x - cur) - 1;
    long long lowest = 2 * go;
    x -= cur + 1;
    while (limInf < limSup) {
      long long ran = ((x) / ((limSup - limInf + 1) / 4ll)) % 4;
      x -= ((limSup - limInf + 1) / 4ll) * ran;
      lowest += ((limSup - limInf + 1) / 4ll) * v[ran];
      limSup = lowest + ((limSup - limInf + 1) / 4ll) - 1;
      limInf = lowest;
    }
    long long B = lowest;
    long long C = A ^ B;
    if (n % 3 == 1)
      cout << A << '\n';
    else if (n % 3 == 2)
      cout << B << '\n';
    else
      cout << C << '\n';
  }
}
