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;
const int N = 1e5 + 10;
const int M = 1e9 + 7;
const long double PI = acos(-1);
long long t, n;
long long a[100], b[5][4][3] = {{{0, 0}, {0, 0}, {0, 0}},
                                {{0, 1}, {1, 0}, {1, 1}},
                                {{1, 0}, {1, 1}, {0, 1}},
                                {{1, 1}, {0, 1}, {1, 0}}};
void Test() {
  cin >> n;
  long long pw = 1, i;
  for (i = 0; i < 70; i++) a[i] = 0;
  while (pw * 3 < n) n -= pw * 3, pw *= 4;
  long long block = n / 3 - (n % 3 == 0);
  long long e = log2(pw) / 2;
  n %= 3;
  --n;
  if (n == -1) n = 2;
  long long t = 0;
  a[1] = block;
  for (i = 1; i <= e; i++) {
    a[i] += t;
    t = a[i] / 4;
    a[i] %= 4;
  }
  a[e + 1] = 1;
  long long nr = 0;
  for (i = 1; i <= e + 1; i++) {
    if (b[a[i]][n][0] == 1) nr += 1ll << (2 * (i - 1) + 1);
    if (b[a[i]][n][1] == 1) nr += 1ll << (2 * (i - 1));
  }
  cout << nr << '\n';
}
signed main() {
  cin >> t;
  while (t--) Test();
  return 0;
}
