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;
void solve() {
  long long n;
  cin >> n;
  n--;
  long long r = n % 3;
  n /= 3;
  long long u = 0;
  while (n >= (1ll << u)) {
    n -= (1ll << u);
    u += 2;
  }
  long long a[3];
  a[0] = (1ll << u);
  a[1] = 2 * a[0];
  a[2] = 3 * a[0];
  long long t = 0;
  while (n) {
    long long kek = n % 4;
    long long y = (1ll << t);
    if (kek == 0) {
    }
    if (kek == 1) {
      a[0] += 1 * y;
      a[1] += 2 * y;
      a[2] += 3 * y;
    } else if (kek == 2) {
      a[0] += 2 * y;
      a[1] += 3 * y;
      a[2] += 1 * y;
    } else if (kek == 3) {
      a[0] += 3 * y;
      a[1] += 1 * y;
      a[2] += 2 * y;
    }
    n /= 4;
    t += 2;
  }
  cout << a[r] << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
}
