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, ... 
from datetime import datetime
from random import randint
import sys

__author__ = 'ratmir'

[t] = [int(x) for x in sys.stdin.readline().split()]


def getFirstInTriple(count):
    st2 = 1
    while st2 < 3 * count:
        st2 <<= 2
    st2 >>= 2
    firstInTriple = st2 + count - ((st2 - 1) / 3) - 1
    return firstInTriple


def getValue(position):
    tripleIndex = 1 + (position - 1) / 3
    firstInTriple = getFirstInTriple(tripleIndex)
    if position % 3 == 1:
        return firstInTriple
    if position % 3 == 2:
        value = 1
        res = 0
        while firstInTriple > 0:
            x = firstInTriple & 3
            if x == 1:
                res += value << 1
            if x == 2:
                res += 3 * value
            if x == 3:
                res += value
            value <<= 2
            firstInTriple >>= 2
        return res
    if position % 3 == 0:
        value = 1
        res = 0
        while firstInTriple > 0:
            x = firstInTriple & 3
            if x == 1:
                res += 3 * value
            if x == 3:
                res += value << 1
            if x == 2:
                res += value
            value <<= 2
            firstInTriple >>= 2
        return res

output = ""




# start_time = datetime.now()
#
# for j in range(0,10):
#     for i in range(0, 100000):
#         getValue(9999999999900001 + i)
#
# print(datetime.now() - start_time)

for i in range(0, t):
    position = int(sys.stdin.readline())
    print getValue(position)
    #output += str(getValue(position)) + "\n"

#print output


