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 arr[4] = {1, 2, 3};
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long nn = (n + 2) / 3;
    long long xx = 1, c = 0;
    while (nn > xx) nn -= xx, xx *= 4, c++;
    long long ad = ((n + 2) % 3 + 1) * xx;
    long long z = nn - 1;
    long long pos = 1, cc = 1;
    for (long long i = 0; i < c; i++) {
      if ((z) % (cc * 4)) ad += pos * arr[((z) % (cc * 4) + (n + 1 % 3)) % 3];
      pos <<= 2;
      z /= 4;
    }
    cout << ad << endl;
  }
  return 0;
}
