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;
long long nu[4][3] = {{0, 0, 0}, {1, 2, 3}, {2, 3, 1}, {3, 1, 2}};
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long ans = 0;
    long long j = (n - 1) % 3;
    n = (n + 2) / 3;
    long long sum = 0;
    long long i = 0;
    for (; i < 60; i += 2) {
      sum |= 1LL << i;
      if (sum >= n) {
        sum ^= 1LL << i;
        n -= sum + 1;
        ans |= nu[1][j] << i;
        break;
      }
    }
    for (i -= 2; i >= 0; i -= 2) ans |= nu[(n >> i) & 3][j] << i;
    cout << ans << endl;
  }
  return 0;
}
