human
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, ... 
Solve the task in PYTHON.
 Can you provide a Python code to find the lexicographically smallest triple of positive integers (a, b, c) such that?
assistant
#Here is Python code to find the lexicographically smallest triple of positive integers (a, b, c) such that 
.

FAST_IO = 1
if FAST_IO:
    import io, sys, atexit
    rr = iter(sys.stdin.read().splitlines()).next
    sys.stdout = _OUTPUT_BUFFER = io.BytesIO()
    @atexit.register
    def write():
        sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
else:
    rr = raw_input
rri = lambda: int(rr())
rrm = lambda: map(int, rr().split())

###
def base4(n):
    ans = []
    while n:
        ans.append(n%4)
        n //= 4
    return ans

def solve(n):
    if n <= 3:
        return n
    n -= 3

    for e in xrange(1, 100):
        base = 4 ** e
        
        if n <= 3 * base:
            root = (n - 1) // 3 + base  # eg. 4, 5, 6, 7
            second = root + base  # 8, 9, 10, 11
            b4 = base4(root-base)
            #print("B", b4)
            for i, x in enumerate(b4):
                if 1 <= x <= 2:
                    second += 4 ** i
                elif x == 3:
                    second -= 2 * (4 ** i)
            third = root ^ second
            if n % 3 == 1:
                return root
            elif n % 3 == 2:
                return second
            else:
                return third
        n -= 3 * base

T = rri()
for _ in xrange(T):
    print solve(rri())