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;
int t;
long long n;
long long f(long long n) {
  long long x = 0, y = 0, z = 0;
  long long cnt = 0;
  long long g = (n - 1) / 3 + 1;
  long long r = (n - 1) % 3;
  long long p4 = 0;
  long long sum = 0;
  while ((1LL << (p4 * 2)) < g) {
    g -= (1LL << (p4 * 2));
    p4++;
  }
  g--;
  x += (1LL << (p4 * 2));
  y += x * 2;
  z += x * 3;
  for (long long b = p4 - 1; b >= 0; b--) {
    long long d = (g >> (b * 2)) % 4;
    if (d == 1) {
      x += (1LL << (b * 2));
      y += (1LL << (b * 2 + 1));
      z += (1LL << (b * 2)) + (1LL << (b * 2 + 1));
    } else if (d == 2) {
      z += (1LL << (b * 2));
      x += (1LL << (b * 2 + 1));
      y += (1LL << (b * 2)) + (1LL << (b * 2 + 1));
    } else if (d == 3) {
      y += (1LL << (b * 2));
      z += (1LL << (b * 2 + 1));
      x += (1LL << (b * 2)) + (1LL << (b * 2 + 1));
    }
  }
  if (r == 0) return x;
  if (r == 1) return y;
  if (r == 2) return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    cout << f(n) << "\n";
  }
  return 0;
}
