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 solve(long long val, long long st) {
  if (st == 1) return 0;
  long long help = st / 4;
  if (val < help) return solve(val, st / 4);
  if (val < 2 * help) return solve(val - help, st / 4) + 2 * help;
  if (val < 3 * help) return solve(val - 2 * help, st / 4) + 3 * help;
  return solve(val - 3 * help, st / 4) + help;
}
long long solve_third(long long val, long long st) {
  if (st == 1) return 0;
  long long help = st / 4;
  if (val < help) return solve_third(val, st / 4);
  if (val < 2 * help) return solve_third(val - help, st / 4) + 3 * help;
  if (val < 3 * help) return solve_third(val - 2 * help, st / 4) + help;
  return solve_third(val - 3 * help, st / 4) + 2 * help;
}
long long query(long long n) {
  if (n % 3 == 1) {
    long long mx = 1;
    while (4 * mx <= n) mx = mx * 4;
    return mx + (n - mx) / 3;
  }
  if (n % 3 == 2) {
    long long mx = 1;
    while (4 * mx <= n) mx = mx * 4;
    return 2 * mx + solve((n - mx) / 3, mx);
  }
  long long mx = 1;
  while (4 * mx <= n) mx = mx * 4;
  return 3 * mx + solve_third((n - mx) / 3, mx);
}
int main() {
  int t;
  scanf("%i", &t);
  for (int i = 1; i <= t; i++) {
    long long n;
    scanf("%lld", &n);
    printf("%lld\n", query(n));
  }
  return 0;
}
