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 next_val[] = {0, 2, 3, 1};
vector<long long> v;
long long find_se(long long base_value, long long pos, long long width) {
  if (width == 0) {
    return base_value;
  }
  return find_se(base_value + width * next_val[pos / width], pos % width,
                 width / 4);
}
int main() {
  long long t;
  long long n;
  cin >> t;
  while (t--) {
    cin >> n;
    long long base = (n - 1) / 3;
    long long type = (n - 1) % 3;
    long long i = 0;
    long long cummulative = 0;
    while (base >= (1ll << (2 * i)) + cummulative) {
      cummulative += 1ll << (2 * i);
      i++;
    }
    long long relative_base = base - cummulative;
    long long base_value = 1ll << (2 * i);
    long long fe = relative_base + base_value;
    long long se = 0;
    if (type == 0) {
      cout << fe;
    } else {
      se = find_se(base_value << 1ll, relative_base, base_value / 4);
      if (type == 1) {
        cout << se;
      } else {
        cout << (fe ^ se);
      }
    }
    cout << "\n";
  }
}
