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, ... 
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
from itertools import combinations

t = int(input())
for tt in range(t):
    n = int(input()) - 1
    mod3 = n % 3
    n = n // 3
    i = 0
    while n >= 4**i:
        n -= 4**i
        i += 1
    if i == 0:
        a = 1; b = 2; c = 3
    else:
        a = 4**i + n
        left = 0
        right = 4**i
        b = 4**i * 2
        t = n
        while right - left > 1:
            mid1 = (left * 3 + right) // 4
            mid2 = (left * 2 + right * 2) // 4
            mid3 = (left + right * 3) // 4
            rng = right - left
            if left <= t < mid1:
                b += 0
            elif mid1 <= t < mid2:
                b += rng // 4 * 2
            elif mid2 <= t < mid3:
                b += rng // 4 * 3
            else:
                b += rng // 4 * 1
            t %= rng // 4 
            right //= 4
        c = a ^ b
    # print("#", tt + 1)
    if mod3 == 0:
        print(a)
    elif mod3 == 1:
        print(b)
    else:
        print(c)

# not_used = set([item for item in range(1, 200)])
# for i in range(25):
#     li = list(not_used)
#     li.sort()
#     for comb in combinations(li, 3):
#         if comb[0] ^ comb[1] ^ comb[2] == 0:
#             print(comb)
#             not_used.remove(comb[0])
#             not_used.remove(comb[1])
#             not_used.remove(comb[2])
#             break