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 long long int conv[][3] = {{1, 2, 3}, {2, 3, 1}, {3, 1, 2}};
void solve(long long int n) {
  int r = n % 3;
  if (r != 0) n = (n / 3 + 1) * 3;
  if (r == 0)
    r = 2;
  else
    r = r - 1;
  int k = 0;
  long long int cnt = 3;
  while (cnt < n) {
    k++;
    cnt += 3 * pow(4, k);
  }
  cnt -= 3 * pow(4, k);
  cnt = n - cnt;
  int i = 0;
  long long int ans = 0;
  ans += conv[0][r] << (2 * (k - i));
  int j;
  for (i = 1; i < k + 1; i++) {
    for (j = 0; j < 4; j++) {
      cnt -= 3 * pow(4, k - i);
      if (cnt <= 0) {
        cnt += 3 * pow(4, k - i);
        break;
      }
    }
    if (j > 0) ans += conv[j - 1][r] << (2 * (k - i));
  }
  printf("%lld\n", ans);
}
int main() {
  int t;
  long long int n;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld", &n);
    solve(n);
  }
  return 0;
}
