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;
unsigned long long next(unsigned long long n) {
  unsigned long long mask = 3;
  int k = 0;
  unsigned long long ret = 0;
  while (mask <= n || (mask & n) != 0) {
    unsigned long long cur_ret = n & mask;
    if (cur_ret != 0) {
      cur_ret += (1LL << k);
      cur_ret &= mask;
      if (cur_ret == 0) cur_ret = (1LL << k);
    }
    ret |= cur_ret;
    mask <<= 2;
    k += 2;
  }
  return ret;
}
unsigned long long get_num_1(unsigned long long n) {
  unsigned long long base = 1;
  while ((base << 2) <= n) {
    base <<= 2;
  }
  n -= (base - 1);
  return base + n / 3;
}
unsigned long long get_num_2(unsigned long long n) {
  unsigned long long base = get_num_1(n - 1);
  return next(base);
}
unsigned long long get_num_0(unsigned long long n) {
  unsigned long long base = get_num_2(n - 1);
  return next(base);
}
unsigned long long get_num(unsigned long long n) {
  switch (n % 3) {
    case 0:
      return get_num_0(n);
    case 1:
      return get_num_1(n);
    case 2:
      return get_num_2(n);
    default:
      return 0;
  }
}
int main() {
  int num;
  cin >> num;
  for (int i = 0; i < num; ++i) {
    unsigned long long n;
    cin >> n;
    unsigned long long num1 = get_num(n);
    cout << num1 << endl;
  }
  return 0;
}
