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;
long long n;
vector<long long> b = {0, 2, 3, 1};
pair<long long, long long> dfs(long long v) {
  if (v == 0) return {4, 8};
  if (v == 1) return {5, 10};
  if (v == 2) return {6, 11};
  if (v == 3) return {7, 9};
  pair<long long, long long> p = dfs(v / 4 - 1);
  p.first <<= 2LL;
  p.second <<= 2LL;
  p.first |= v % 4;
  p.second |= b[v % 4];
  return p;
}
void solve() {
  if (n < 3) {
    cout << n + 1 << '\n';
    return;
  }
  long long g = n / 3 - 1;
  pair<long long, long long> p = dfs(g);
  vector<long long> ans = {p.first, p.second, (p.first ^ p.second)};
  cout << ans[n % 3] << '\n';
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  for (; t--;) {
    cin >> n;
    --n;
    solve();
  }
}
