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;
namespace fio {
streambuf* in = cin.rdbuf();
char bb[1000000], *s = bb, *t = bb;
inline long long read() {
  long long x = 0;
  char ch =
      (s == t && (t = (s = bb) + in->sgetn(bb, 1000000), s == t) ? EOF : *s++);
  while (ch < 48)
    ch = (s == t && (t = (s = bb) + in->sgetn(bb, 1000000), s == t) ? EOF
                                                                    : *s++);
  while (ch >= 48)
    x = x * 10 + ch - 48,
    ch = (s == t && (t = (s = bb) + in->sgetn(bb, 1000000), s == t) ? EOF
                                                                    : *s++);
  return x;
}
}  // namespace fio
using fio::read;
int n, x, y;
int sxx(int x, int y) {
  if (y == 1) {
    if (x == 0) return 0;
    if (x == 1) return 1;
    if (x == 2) return 2;
    if (x == 3) return 3;
  }
  if (y == 2) {
    if (x == 0) return 0;
    if (x == 1) return 2;
    if (x == 2) return 3;
    if (x == 3) return 1;
  }
  if (y == 3) {
    if (x == 0) return 0;
    if (x == 1) return 3;
    if (x == 2) return 1;
    if (x == 3) return 2;
  }
  return -1;
}
long long dfs(long long n) {
  if (n <= 3) return n;
  long long x = (n - 1) / 3 - 1, y = (n % 3) == 0 ? 3 : (n % 3);
  if (y == 1)
    return (dfs((x >> 2) * 3 + y) << 2) | sxx(x & 3, 1);
  else if (y == 2)
    return (dfs((x >> 2) * 3 + y) << 2) | sxx(x & 3, 2);
  return (dfs((x >> 2) * 3 + y) << 2) | sxx(x & 3, 3);
}
int main() {
  int T = read();
  while (T--) cout << dfs(read()) << endl;
}
