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>
struct result {
  long long level;
  long long sum_up_to;
};
result find_level(long long n) {
  result res;
  res.sum_up_to = 0;
  for (long long level = 0; level <= 100; level += 2) {
    long long this_level = (1ll << level);
    if (n < res.sum_up_to + this_level) {
      res.level = level;
      return res;
    }
    res.sum_up_to += this_level;
  }
  return res;
}
long long solve1(long long n) {
  auto lvl = find_level(n);
  return (n - lvl.sum_up_to) + (1ll << lvl.level);
}
long long solve2(long long n) {
  static const long long mp[] = {0, 2, 3, 1};
  auto lvl = find_level(n);
  long long offset = (n - lvl.sum_up_to);
  long long base = ((1ll << lvl.level) << 1);
  long long spec_offset = 0;
  for (long long level = lvl.level - 2; level >= 0; level -= 2) {
    long long bitmask = 3 * (1ll << level);
    long long idx = (offset & bitmask) >> level;
    spec_offset += mp[idx] * (1ll << level);
  }
  return base + spec_offset;
}
long long solve3(long long n) {
  static const long long mp[] = {0, 3, 1, 2};
  auto lvl = find_level(n);
  long long offset = (n - lvl.sum_up_to);
  long long base = ((1ll << lvl.level) << 1) + ((1ll << lvl.level));
  long long spec_offset = 0;
  for (long long level = lvl.level - 2; level >= 0; level -= 2) {
    long long bitmask = 3 * (1ll << level);
    long long idx = (offset & bitmask) >> level;
    spec_offset += mp[idx] * (1ll << level);
  }
  return base + spec_offset;
}
void solve() {
  long long n;
  std::cin >> n;
  long long ans = 0;
  if (n % 3 == 1) {
    ans = solve1((n - 1) / 3);
  } else if (n % 3 == 2) {
    ans = solve2((n - 2) / 3);
  } else if (n % 3 == 0) {
    ans = solve3((n - 3) / 3);
  }
  std::cout << ans << "\n";
}
int main() {
  std::ios::sync_with_stdio(false);
  int t;
  std::cin >> t;
  while (t--) solve();
  return 0;
}
