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 test;
long long n;
long long cal(long long a) {
  long long ans = 0;
  for (int i = 0; i < 61; i += 2) {
    long long cyc = 1LL << i;
    if (a < cyc - 1) {
      continue;
    }
    vector<long long> d = {1, 2, 3};
    long long x = a - cyc + 1;
    if (x < cyc * 3) {
      ans |= d[x % 3] << i;
      continue;
    }
    x -= cyc * 3;
    if (x < cyc * 3) {
      continue;
    }
    x -= cyc * 3;
    x %= cyc * 3 * 4;
    if (x >= cyc * 3 * 3) {
      continue;
    }
    while (x >= cyc * 3) {
      x -= cyc * 3;
      rotate(d.begin(), d.begin() + 1, d.end());
    }
    if (x <= cyc * 3) {
      ans |= d[x % 3] << i;
      continue;
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> test;
  while (test--) {
    cin >> n;
    cout << cal(n - 1) << '\n';
  }
  return 0;
}
