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;
function<void(void)> ____ = []() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
};
using LL = int_fast64_t;
void solve() {
  LL n;
  cin >> n;
  LL m = (n - 1) / 3 + 1, md = n % 3 ? n % 3 : 3, ret = 0, bit = 1, num = 0,
     a[4];
  if (md == 1)
    a[0] = 3, a[1] = 0, a[2] = 1, a[3] = 2;
  else if (md == 2)
    a[0] = 1, a[1] = 0, a[2] = 2, a[3] = 3;
  else
    a[0] = 2, a[1] = 0, a[2] = 3, a[3] = 1;
  while (true) {
    if (bit >= m) {
      ret += (md << num);
      break;
    }
    m -= bit;
    ret += (a[((m - 1) / bit + 1) % 4] << num);
    bit <<= 2, num += 2;
  }
  cout << ret << endl;
}
int main() {
  ____();
  int T;
  for (cin >> T; T; T--) solve();
  return 0;
}
