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, ... 
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author kamel
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        long[] data = {1, 2, 3};
        long[] power = new long[90];
        String[] twelveBit = {"0", "0", "0", "1", "2", "3", "2", "3", "1", "3", "1", "2"};

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            power[0] = 1L;
            for (int i = 1; i < power.length; i++) {
                power[i] = power[i - 1] * 4L;
            }
            int cases = in.nextInt();
            for (int i = 0; i < cases; i++) {
                long N = in.nextLong();
                out.println(Long.valueOf(f(N), 4));
            }
        }

        private String f(long n) {
            //System.out.println(n);
            if (n <= 3) {
                return String.valueOf(n);
            }
            int currentBit = 2;
            long nextBit = currentBit + 2;
            while ((1L << nextBit) <= n) {
                nextBit += 2;
            }

            long left = n - (1L << (nextBit - 2)) + 1;
            long firstPart = data[(int) ((left - 1L) % 3)];
            return firstPart + f2(left - 1, (int) nextBit / 2 - 1);
        }

        private String f2(long left, int bit) {
            //System.out.println(String.format("%d %d", left, bit));
            if (bit == 1) {
                assert left < twelveBit.length;
                return twelveBit[(int) left];
            }

            long singlePass = 3L * power[(bit - 1)];
            long passes = left / singlePass;
            long singleLeft = left % singlePass;
            return twelveBit[(int) (3 * passes + (singleLeft % 3))] + f2(singleLeft, bit - 1);
        }

    }
}

