1338_C. Perfect Triples
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, ... 
```python
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())))

```