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 int logo(long long int x, long long int b) {
  long long int ans = 0;
  while (x > 0) {
    x /= b;
    ans++;
  }
  return ans;
}
long long int powr(long long int b, long long int x) {
  if (x == 0) return 1;
  long long int sq = powr(b, x / 2);
  if (x & 1) return sq * sq * b;
  return sq * sq;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    long long int n;
    cin >> n;
    long long int gr = (n - 1) / 3;
    long long int sp = (n - 1) % 3;
    long long int ser = logo(3 * gr + 1, 4) - 1;
    long long int disp = gr - (pow(4, ser) - 1) / 3;
    long long int a = (1LL << (2 * ser)) + disp;
    long long int c = (3LL << (2 * ser));
    for (int i = 0; i <= ser; ++i) {
      if (disp % 4 == 0) {
      } else if (disp % 4 == 1) {
        c |= (3LL << (2 * i));
      } else if (disp % 4 == 2) {
        c |= (1LL << (2 * i));
      } else {
        c |= (2LL << (2 * i));
      }
      disp /= 4;
    }
    long long int b = a ^ c;
    if (sp % 3 == 0)
      cout << a << endl;
    else if (sp % 3 == 1)
      cout << b << endl;
    else
      cout << c << endl;
  }
}
