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, ... 
//package javaapplication5;

import java.io.*;
import java.util.*;
 
public class utkarsh {
    
    BufferedReader br;
    PrintWriter out;
    
    long mod = (long) (1e9 + 7), inf = (long) (3e18);
    
    void print(int x) {
        out.print(x +" ->\t");
        for(int i = 9; i >= 0; i--) {
            out.print(((x >> i) & 1) +" ");
            if(i%2 == 0)    out.print(" ");
        }
        out.println();
    }
    
    void brute(int n) {
        boolean vst[] = new boolean[n];
        for(int i = 1; i < n; i++) {
            if(vst[i])  continue;
            for(int j = i+1; j < n; j++) {
                if(vst[j] || vst[i^j])  continue;
                vst[i] = true;  print(i);
                vst[j] = true;  print(j);
                vst[i^j] = true;    print(i^j);
                out.println();
                break;
            }
        }
    }
    
    void solve() {
        //brute(256);
        int t = ni();
        while(t-- > 0) {
            long n = nl();
            long pow2 = 1;
            long exp = 0;
            while(pow2 <= n) {
                pow2 <<= 1; exp++;
            }
            if(exp % 2 == 1) {
                exp++;  pow2 <<= 1;
            }
            long pos = n-1;
            long index = pos % 3;
            
            long ans = (pos % 3) + 1;
            exp -= 2;
            pos -= ((1L << exp) - 1);
            pos /= 3;
            
            while(exp > 0) {
                exp -= 2;
                long sz = (1L << exp);
                ans <<= 2;
                long ind = pos / sz;
                if(ind == 0) {
                } else if(ind == 1) {
                    if(index == 0) {
                        ans |= 1;
                    } else if(index == 1) {
                        ans |= 2;
                    } else {
                        ans |= 3;
                    }
                    pos -= sz;
                } else if(ind == 2) {
                    if(index == 0) {
                        ans |= 2;
                    } else if(index == 1) {
                        ans |= 3;
                    } else {
                        ans |= 1;
                    }
                    pos -= sz * 2;
                } else {
                    if(index == 0) {
                        ans |= 3;
                    } else if(index == 1) {
                        ans |= 1;
                    } else {
                        ans |= 2;
                    }
                    pos -= sz * 3;
                }
            }
            out.println(ans);
        }
    }
    
    long mp(long b, long e) {
        long r = 1;
        while(e > 0) {
            if( (e&1) == 1 )    r = (r * b) % mod;
            b = (b * b) % mod;
            e >>= 1;
        }
        return r;
    }
    
    // -------- I/O Template -------------
    
    char nc() {
        return ns().charAt(0);
    }
    
    String nLine() {
        try {
            return br.readLine();
        } catch(IOException e) {
            return "-1";
        }
    }
    
    double nd() {
        return Double.parseDouble(ns());
    }
    
    long nl() {
        return Long.parseLong(ns());
    }
    
    int ni() {
        return Integer.parseInt(ns());
    }
    
    StringTokenizer ip;
    
    String ns() {
        if(ip == null || !ip.hasMoreTokens()) {
            try {
                ip = new StringTokenizer(br.readLine());
            } catch(IOException e) {
                throw new InputMismatchException();
            }
        }
        return ip.nextToken();
    }
    
    void run() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.flush();
    }
    
    public static void main(String[] args) {
        new utkarsh().run();
    }
}
