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, ... 
mod = 1000000007
eps = 10**-9


def main():
    import sys
    input = sys.stdin.buffer.readline

    for _ in range(int(input())):
        N = int(input())
        if N <= 3:
            print(N)
            continue
        if N%3 == 1:
            k = N.bit_length()-1
            if k&1:
                k -= 1
            print(2**k + (N-2**k) // 3)
        elif N%3 == 2:
            N1 = N-1
            k = N1.bit_length()-1
            if k&1:
                k -= 1
            ans1 = 2**k + (N1-2**k) // 3
            ans = 0
            cnt = 0
            while ans1:
                a = ans1 % 4
                if a == 1:
                    a = 2
                elif a == 2:
                    a = 3
                elif a == 3:
                    a = 1
                ans += a * (4**cnt)
                cnt += 1
                ans1 >>= 2
            print(ans)
        else:
            N1 = N - 2
            k = N1.bit_length()-1
            if k & 1:
                k -= 1
            ans1 = 2 ** k + (N1 - 2 ** k) // 3
            ans = 0
            cnt = 0
            while ans1:
                a = ans1 % 4
                if a == 1:
                    a = 3
                elif a == 2:
                    a = 1
                elif a == 3:
                    a = 2
                ans += a * (4 ** cnt)
                cnt += 1
                ans1 >>= 2
            print(ans)


if __name__ == '__main__':
    main()
