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 main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long p4 = 1;
    while (p4 * 4 <= n) p4 *= 4;
    long long k = p4 + (n - p4) / 3;
    if (n % 3 == 1) {
      cout << k << endl;
    } else if (n % 3 == 2) {
      long long ans = 2;
      k ^= p4;
      while (p4 > 1) {
        ans *= 4;
        p4 /= 4;
        long long x = k / p4;
        k %= p4;
        switch (x) {
          case 0:
            break;
          case 1:
            ans += 2;
            break;
          case 2:
            ans += 3;
            break;
          case 3:
            ans += 1;
            break;
        }
      }
      cout << ans << '\n';
    } else {
      long long ans = 3;
      k ^= p4;
      while (p4 > 1) {
        ans *= 4;
        p4 /= 4;
        long long x = k / p4;
        k %= p4;
        switch (x) {
          case 0:
            break;
          case 1:
            ans += 3;
            break;
          case 2:
            ans += 1;
            break;
          case 3:
            ans += 2;
            break;
        }
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
