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 T;
long long n;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%lld", &n);
    long long p = (n - 1) / 3, s = 1, all = 0;
    while (p >= s) p -= s, s <<= 2;
    if (n % 3 == 1)
      printf("%lld\n", s + p);
    else if (n % 3 == 2) {
      long long ans = s << 1, c = 1;
      while (p) {
        if (p % 4 == 1)
          ans += c * 2;
        else if (p % 4 == 2)
          ans += c * 3;
        else if (p % 4 == 3)
          ans += c;
        p >>= 2, c <<= 2;
      }
      printf("%lld\n", ans);
    } else {
      long long ans = s * 3, c = 1;
      while (p) {
        if (p % 4 == 1)
          ans += c * 3;
        else if (p % 4 == 2)
          ans += c * 1;
        else if (p % 4 == 3)
          ans += c * 2;
        p >>= 2, c <<= 2;
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}
