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.util.Scanner;
public class Solution {
    static long solve1(long n)
    {
        long a=1;
        while(n>=a)a*=4;
        a/=4;
        n-=a;
        n/=3;
        return a+n;
    }
    static long quad_search(long l,long r,long n)
    {
        if(l==r)return l;
        long cnt=(r-l+1)/4;
        long part=n/cnt;
        if(part==0)return quad_search(l,l+cnt-1,n);
        else if(part==3)return quad_search(l+cnt,l+2*cnt-1,n-part*cnt);
        else if(part==1)return quad_search(l+2*cnt,l+3*cnt-1,n-part*cnt);
        return quad_search(l+3*cnt,r,n-part*cnt);
    }
    static long solve2(long n)
    {
        long a=1;
        while(n>=a)a*=4;
        n-=a/4;
        n/=3;
        long l=a/2,r=a/2+a/4-1;
        return quad_search(l,r,n);
    }
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        long t=sc.nextLong();
        while(t-->0)
        {
            long n=sc.nextLong();
            if(n%3==1)System.out.println(solve1(n));
            else if(n%3==2)System.out.println(solve2(n));
            else System.out.println(solve1(n-2)^solve2(n-1));
        }
    }
}
