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;
template <class K>
inline bool cmax(K& a, const K& b) {
  return (a < b) ? a = b, 1 : 0;
}
template <class K>
inline bool cmin(K& a, const K& b) {
  return (a > b) ? a = b, 1 : 0;
}
inline long long read() {
  register long long s = 0;
  register char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) s = (s << 1) + (s << 3) + c - '0', c = getchar();
  return s;
}
int T, c;
long long n, f[31], A, B, m, x, base, res;
int main() {
  register int i;
  T = read(), f[0] = 1;
  for (i = 1; i < 30; i++) f[i] = f[i - 1] * 4;
  while (T--) {
    n = read();
    c = lower_bound(f, f + 30, n + 1) - f - 1;
    n -= f[c] - 1, m = (n - 1) / 3, A = f[c] + m, base = 1;
    B = 0, res = A;
    while (A) {
      switch (A % 4) {
        case 1:
          B += 2 * base;
          break;
        case 2:
          B += 3 * base;
          break;
        case 3:
          B += base;
          break;
      }
      base = base * 4, A /= 4;
    }
    A = res;
    if (n % 3 == 1)
      printf("%lld\n", A);
    else if (n % 3 == 2)
      printf("%lld\n", B);
    else
      printf("%lld\n", A ^ B);
  }
  return 0;
}
