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 main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n = 500;
  vector<int> res;
  set<int> s;
  for (int i = 0; i < 1000 * n; ++i) s.insert(i + 1);
  while (res.size() < n) {
    int a = *s.begin();
    int b, c;
    s.erase(s.begin());
    for (int i : s) {
      if (s.find(a ^ i) != s.end()) {
        b = i, c = a ^ b;
        break;
      }
    }
    s.erase(b);
    s.erase(c);
    res.push_back(a);
    res.push_back(b);
    res.push_back(c);
  }
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    if (n < 100) {
      cout << res[n - 1] << "\n";
      continue;
    }
    auto first = [](long long n) {
      long long x = 1;
      while (x * 4 <= n) x *= 4;
      return x + (n - x) / 3;
    };
    auto secsec = [](long long ind) {
      int d[] = {0, 2, 3, 1, 8, 10, 11, 9, 12, 14, 15, 13, 4, 6, 7, 5};
      return d[ind];
    };
    auto second = [secsec](long long n) {
      long long x = 1;
      while (x * 4 <= n) x *= 4;
      long long ind = (n - x) / 3;
      long long t = 1LL << 60;
      long long res = x * 2;
      while (t > 0) {
        res += t * secsec(ind / t), ind = ind % t;
        t /= 16;
      }
      return res;
    };
    if (n % 3 == 1) {
      cout << first(n) << "\n";
    } else if (n % 3 == 2) {
      cout << second(n) << "\n";
    } else {
      cout << (first(n) ^ second(n)) << "\n";
    }
  }
  return 0;
}
