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 pow4[31];
long long psa[32];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  pow4[0] = 1;
  psa[0] = 0;
  psa[1] = 1;
  for (int i = 1; i < 31; i++) {
    pow4[i] = pow4[i - 1] * 4;
    psa[i + 1] = psa[i] + pow4[i];
  }
  int tt;
  cin >> tt;
  while (tt--) {
    long long n;
    cin >> n;
    if (n <= 3) {
      cout << n << '\n';
      continue;
    }
    n -= 3;
    long long m = (n + 2) / 3;
    int e = 31;
    while (m < psa[e]) {
      e--;
    }
    m -= psa[e];
    long long x = pow4[e] + m;
    long long y = pow4[e] * 2;
    for (int i = 0; i < e; i++) {
      int b = (x >> (2 * i)) % 4;
      if (b == 1) {
        y |= 2LL << (2 * i);
      } else if (b == 2) {
        y |= 3LL << (2 * i);
      } else if (b == 3) {
        y |= 1LL << (2 * i);
      }
    }
    if (n % 3 == 1) {
      cout << x << '\n';
    } else if (n % 3 == 2) {
      cout << y << '\n';
    } else {
      cout << (x ^ y) << '\n';
    }
  }
  return 0;
}
