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 main() {
  long long a[200000];
  long long b[200000];
  long long c[200000];
  for (long long i = 0; i < (long long)(200000); i++) a[i] = 0;
  long long n, k, t;
  long long x, y, z;
  long long xx, yy, zz;
  long long s, ans;
  bool v = true;
  ans = 0;
  a[0] = 1;
  for (long long i = 0; i < (long long)(30); i++) a[i + 1] = a[i] * 4;
  scanf("%lld", &t);
  for (long long tt = 0; tt < (long long)(t); tt++) {
    scanf("%lld", &n);
    k = (n + 2) % 3;
    n = (n + 2) / 3;
    x = 0;
    y = 0;
    while (true) {
      y += a[x];
      if (n <= y) break;
      x++;
    }
    xx = a[x] * 2 - (y - n) - 1;
    z = a[x] - (y - n) - 1;
    yy = 2 * a[x];
    for (long long i = 0; i < (long long)(x); i++) {
      c[i] = z % 4;
      z /= 4;
      if (c[i] == 1)
        c[i] = 2;
      else if (c[i] == 2)
        c[i] = 3;
      else if (c[i] == 3)
        c[i] = 1;
      yy += (c[i] * a[i]);
    }
    zz = xx ^ yy;
    if (k == 0)
      printf("%lld\n", xx);
    else if (k == 1)
      printf("%lld\n", yy);
    else if (k == 2)
      printf("%lld\n", zz);
  }
  return 0;
}
