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, ... 
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
If I'm the sun, you're the moon
Because when I go up, you go down
*******************************
I'm working for the day I will surpass you
https://www.a2oj.com/Ladder16.html
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class x1339E
   {
      public static void main(String omkar[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int T = Integer.parseInt(st.nextToken());
         StringBuilder sb = new StringBuilder();
         int[] akame = {0, 1, 2, 3, 4, 8, 12, 5, 10, 15, 6, 11, 13, 7, 9, 14};
         while(T-->0)
         {
            long N = Long.parseLong(infile.readLine());
            if(N < 16)
            {
               sb.append(akame[(int)N]+"\n");
               continue;
            }
            long pow4 = 4L;
            int cnt = 1;
            while(pow4 <= N)
            {
               pow4 *= 4;
               cnt++;
            }
            pow4 /= 4;
            cnt--;
            //4 = 0,0,0 
            //5 = 1,2,3
            //6 = 2,3,1
            //7 = 3,1,2
            int[] num = new int[cnt];
            long before = pow4-1;
            for(int i=0; i < cnt; i++)
            {
               num[i] = 4;
               long temp = 3L;
               for(int qw=i+1; qw < cnt; qw++)
                  temp *= 4;
               long cons = 1L;
               while(temp*cons+before < N)
               {
                  cons++;
                  num[i]++;
               }
               before += temp*(cons-1);
            }
            int[][] map = {{0,0,0}, 
                           {1,2,3},
                           {2,3,1},
                           {3,1,2}};
            String res = (N%3)+"";
            if(N%3 == 0)   res = "3";
            StringBuilder sb1 = new StringBuilder(res);
            for(int i=0; i < cnt; i++)
            {
               int val = map[num[i]-4][(int)((N-1)%3)];
               sb1.append(val);
            }
            sb.append(Long.parseLong(sb1.toString(), 4)+"\n");
         }
         System.out.print(sb);
      }
   }