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;
int64_t trans[3][4] = {
    {0, 1, 2, 3},
    {0, 2, 3, 1},
    {0, 3, 1, 2},
};
int64_t solve(int64_t N) {
  N--;
  int64_t tripNum = N / 3;
  int numDig = 0;
  while (tripNum >= (int64_t(1) << (2 * numDig))) {
    tripNum -= (int64_t(1) << (2 * numDig));
    numDig++;
  }
  int64_t v = (int64_t(1) << (2 * numDig)) + tripNum;
  int64_t ans = 0;
  for (int d = 0; d <= numDig; d++) {
    ans |= trans[N % 3][(v >> (2 * d)) & 3] << (2 * d);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    int64_t N;
    cin >> N;
    int64_t ans = solve(N);
    cout << ans << '\n';
  }
  return 0;
}
