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 maxn = (int)1e6 + 100;
const int mod = (int)1e9 + 7;
long long ans[5];
void solve() {
  long long n;
  scanf("%lld", &n);
  if (n <= 3) return (void)printf("%lld\n", n);
  long long N = (n - 1) / 3 + 1, b = 1;
  for (; b < N; b *= 4) N -= b;
  ans[1] = N + b - 1, ans[2] = b * 2;
  long long sz = b;
  while (sz > 1) {
    sz /= 4;
    if ((N - 1) / sz == 1)
      N -= sz, ans[2] += sz * 2;
    else if ((N - 1) / sz == 2)
      N -= sz * 2, ans[2] += sz * 3;
    else if ((N - 1) / sz == 3)
      N -= sz * 3, ans[2] += sz;
  }
  ans[0] = ans[1] ^ ans[2];
  printf("%lld\n", ans[n % 3]);
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
}
