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 long long one = 1;
long long mp[] = {0, 2, 3, 1};
long long solve(long long n) {
  if (n < 4) return n;
  long long _n = n;
  n -= 4;
  long long a, b, c;
  for (int i = 2;; i += 2) {
    long long can = 3 * (one << i);
    if (n < can) {
      a = (one << i) | (n / 3);
      b = one << (i + 1);
      n /= 3;
      for (int j = i - 2; j >= 0; j -= 2) {
        long long x = n / (one << j);
        b |= mp[x] << j;
        n -= x * (one << j);
      }
      break;
    }
    n -= can;
  }
  assert(n == 0);
  c = a ^ b;
  if (_n % 3 == 1) return a;
  if (_n % 3 == 2) return b;
  return c;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cout << solve(n) << "\n";
  }
  return 0;
}
