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;
const int N = 2e5 + 5;
int t;
long long n, ans[5];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    if (n <= 3) {
      cout << n << "\n\n";
      continue;
    }
    long long res = n / 3;
    if (n % 3 == 0) res--;
    long long now = 1, base = 1, cnt = 0;
    while (now <= res) {
      cnt++;
      base *= 4;
      now += base;
    }
    now -= base;
    res -= now;
    now = 1;
    ans[1] = ans[2] = ans[3] = 0;
    for (int i = 1; i <= cnt; i++) {
      long long j = res % 4;
      if (j == 1) {
        ans[1] += now * 1;
        ans[2] += now * 2;
        ans[3] += now * 3;
      }
      if (j == 2) {
        ans[1] += now * 2;
        ans[2] += now * 3;
        ans[3] += now * 1;
      }
      if (j == 3) {
        ans[1] += now * 3;
        ans[2] += now * 1;
        ans[3] += now * 2;
      }
      now *= 4;
      res /= 4;
    }
    long long hah;
    if (n % 3 == 0) {
      hah = 3;
    } else {
      hah = n % 3;
    }
    cout << hah * now + ans[hah] << "\n";
    cout << "\n";
  }
}
