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;
long long pot[30];
int main(void) {
  int t;
  scanf("%d", &t);
  pot[0] = 1;
  for (int i = 1; i <= 27; i++) pot[i] = pot[i - 1] * 4LL;
  while (t--) {
    long long n;
    scanf("%lld", &n);
    int p;
    long long a, b;
    for (int i = 0; i <= 27; i++) {
      if (pot[i] - 1 >= n) {
        p = i - 1;
        a = pot[i - 1], b = 2LL * pot[i - 1];
        break;
      }
    }
    long long grupo = (n - a);
    a += grupo / 3LL;
    p *= 2;
    for (int i = 0; i < p; i += 2) {
      long long b0, b1;
      b0 = (a & (1LL << i));
      b1 = (a & (1LL << (i + 1)));
      if (b0 && !b1)
        b += (1LL << (i + 1));
      else if (!b0 && b1)
        b += (1LL << i), b += (1LL << (i + 1));
      else if (b0 && b1)
        b += (1LL << i);
    }
    if (grupo % 3 == 0)
      printf("%lld\n", a);
    else if (grupo % 3 == 1)
      printf("%lld\n", b);
    else
      printf("%lld\n", a ^ b);
  }
}
