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;
bool sort_func(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first < b.first;
}
long long f(long long a) {
  if (a == 0) return 0;
  if (a == 1) return 2;
  if (a == 2) return 3;
  if (a == 3) return 1;
  while (true) {
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t;
  cin >> t;
  for (long long idx = 0; idx < t; idx++) {
    long long n;
    cin >> n;
    long long nm = n - (n - 1) % 3;
    long long pc = 0, p4 = 1;
    while (n >= p4 * 4) {
      pc++;
      p4 *= 4;
    }
    long long m = (nm - p4) / 3 + p4;
    if (n % 3 == 1)
      cout << m << endl;
    else {
      long long m2 = 0;
      long long p = 1;
      for (long long i = 0; i < pc; i++) {
        m2 += f((m / p) % 4) * p;
        p *= 4;
      }
      m2 += 2 * p4;
      if (n % 3 == 2)
        cout << m2 << endl;
      else
        cout << (m ^ m2) << endl;
    }
  }
  long long _;
  cin >> _;
}
