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>
int main() {
  int t, p;
  scanf("%d", &t);
  for (p = 1; p <= t; p++) {
    long long n;
    scanf("%lld", &n);
    if (n <= 3) {
      printf("%lld\n", n);
      continue;
    }
    long long m = n;
    long long base = 1;
    while (m > 0) {
      m = m / 4;
      base = base * 4;
    }
    base = base / 4;
    if (n % 3 == 1) {
      printf("%lld\n", base + (n - base) / 3);
      continue;
    }
    long long fi = base + (n - base) / 3;
    long long se = base * 2;
    m = (n - base) / 3;
    long long cur = 1;
    while (m > 0) {
      if (m % 4 == 1) se += 2 * cur;
      if (m % 4 == 2) se += 3 * cur;
      if (m % 4 == 3) se += cur;
      m = m / 4;
      cur = cur * 4;
    }
    if (n % 3 == 2)
      printf("%lld\n", se);
    else
      printf("%lld\n", fi ^ se);
  }
  return 0;
}
