Consider the infinite sequence $s$ of positive integers, created by repeating the following steps:

  Find the lexicographically smallest triple of positive integers $(a, b, c)$ such that   $a \oplus b \oplus c = 0$, where $\oplus$ denotes the bitwise XOR operation.  $a$, $b$, $c$ 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 $[a_2, b_2, c_2]$.  Append $a$, $b$, $c$ to $s$ in this order.  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 $t$ ($1 \le t \le 10^5$) — the number of test cases.

Each of the next $t$ lines contains a single integer $n$ ($1\le n \le 10^{16}$) — 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, \dots $

Difficulty: competition

Starter code:

Solutions:
import sys
input = sys.stdin.readline
out = []

t = int(input())
for _ in range(t):
    n = int(input())
    n -= 1
    rem = n % 3
    n //= 3
    
    s = []
    if n:
        n -= 1
        while n >= 0:
            s.append([['00','00','00'],['01','10','11'],['10','11','01'],['11','01','10']][n % 4][rem])
            n //= 4
            n -= 1

    s.append(['1','10','11'][rem])

    s.reverse()

    out.append(int(''.join(s),2))
print('\n'.join(map(str,out)))
    


import sys
input = sys.stdin.readline
t = int(input())
s = [int(input()) for i in range(t)]
res = []
for num in s:
    num -= 1
    mod = num % 3
    num = num // 3
    div = 1
    while True:
        if num // div != 0:
            num -= div
            div *= 4
        else:
            break
    a = div + num
    b = a * 2
    tmp = a
    coff = 1
    while True:
        if tmp == 0:
            break
        if tmp % 4 == 2:
            b -= coff
        if tmp % 4 == 3:
            b -= coff * 5
        tmp = tmp // 4
        coff *= 4
    c = a ^ b
    if mod == 0:
        print(a)
    if mod == 1:
        print(b)
    if mod == 2:
        print(c)

import sys
input = sys.stdin.readline
import bisect

L=[4**i for i in range(30)]

def calc(x):
    fami=bisect.bisect_right(L,x)-1
    base=(x-L[fami])//3
    cl=x%3

    if cl==1:
        return L[fami]+base
    else:
        cl1=L[fami]+base
        ANS=0

        for i in range(30):
            x=cl1//L[i]%4

            if x==1:
                ANS+=2*L[i]
            elif x==2:
                ANS+=3*L[i]
            elif x==3:
                ANS+=L[i]

        if cl==2:
            return ANS
        else:
            return ANS^cl1

t=int(input())
for tests in range(t):
    print(calc(int(input())))