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;
template <typename T1, typename T2>
inline void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
using ll = long long;
using ld = long double;
const string FILENAME = "input";
void solve() {
  ll n;
  cin >> n;
  n--;
  ll id = n / 3;
  ll pos;
  int it = 0;
  for (ll i = 1;; i *= 4) {
    it++;
    if (i > id) {
      pos = i;
      break;
    }
    id -= i;
  }
  ll a = pos + id;
  ll b = pos * 2;
  ll g[4] = {0, 2, 3, 1};
  for (int i = 0; i < it - 1; i++) {
    pos /= 4;
    ll val = id / pos;
    b += pos * g[val];
    id %= pos;
  }
  if (n % 3 == 0) {
    cout << a;
  } else if (n % 3 == 1) {
    cout << b;
  } else {
    cout << (a ^ b);
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
