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, stdout

t = input()
inp = stdin.readlines()
out = []

def solve(n):
    if n <= 3: return n
    
    tnum = 0
    # find the triplet number first
    if n % 3 == 0:
        tnum = n // 3
    elif n % 3 == 1:
        tnum = (n - 1) // 3 + 1
    else:
        tnum = (n - 2) // 3 + 1

    # find the group number
    cumsum = 0
    p4 = 1
    while cumsum < tnum:
        cumsum += p4
        p4 <<= 2
    p4 >>= 2

    # using the group number, we can determine a
    relpos = tnum - (cumsum - p4)
    a = p4 + relpos - 1
    
    # now we need to determine b
    size = p4 # size of the group

    add = 0

    cur = p4 # current size
    lo = p4
    while cur > 1:
        sz = (cur >> 2)

        pos = -1
        for i in xrange(4):
            if lo + i * sz <= a and a < lo + (i + 1) * sz:
                pos = i
                break

        cur = sz
        lo = lo + pos * sz

        if pos == 0: pass
        elif pos == 1: add += 2 * sz
        elif pos == 2: add += 3 * sz
        else: add += sz

    b = p4 + p4 + add

    c = a ^ b

    if n % 3 == 1: return a
    elif n % 3 == 2: return b
    else: return c

for line in inp:
    n = int(line)
    ans = solve(n)
    out.append(str(ans))

stdout.write('\n'.join(out))
