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 sys import stdin
input = stdin.readline
pot = [1] * 40
for i in range(1,40):
	pot[i] = 4 * pot[i-1]
poczatki = [1,2]
for i in range(38):
	os = poczatki[-1]
	poczatki.append(os + pot[i+1])
tab = [[0,1,2,3],[0,2,3,1],[0,3,1,2]]
def odp(ind, m3):
	if ind == 1:
		return (m3+1)
	else:
		m = ind
		while m%4 != 2:
			m -= 1
		i = 0
		while poczatki[i+1] <= m:
			i += 1
		#poczatki[i] to poczatek segu dlugosci pot[i]
		miejsce = (m-poczatki[i])//4
		pre = poczatki[i-1] + miejsce
		return (4*odp(pre,m3) + tab[m3][ind - m])

t = int(input())
for _ in range(t):
    n = int(input())
    a = (n-1)%3
    b = (n-1)//3 + 1
    print(odp(b,a))

