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, ... 
import java.util.*;
import java.io.*;

public class PerfectTriples {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		int t = Integer.parseInt(br.readLine());
		StringBuilder sb = new StringBuilder();

		while (t-- > 0) {
			long n = Long.parseLong(br.readLine());
			sb.append(getElement(n));
			sb.append('\n');
		}

		br.close();

		System.out.print(sb.toString());
	}

	static long getElement(long n) {
		int bits = 1;
		long val = 2;

		while (val <= n) {
			bits++;
			val <<= 1;
		}
		if ((bits & 1) != 0) {
			bits--;
			val >>= 1;
		} else {
			bits -= 2;
			val >>= 2;
		}
		long a = val + (n - val) / 3;
		int mod = (int)((n - val) % 3);
		if (mod == 0) {
			return a;
		}

		long res = 0;
		int position = 0;

		while (position <= bits + 1) {
			int pair = (int)(3 & a);
			long digits;
			if (mod == 1) {
				if (pair == 0) {
					digits = 0;
				} else if (pair == 1) {
					digits = 2;
				} else if (pair == 2) {
					digits = 3;
				} else {
					digits = 1;
				}
				res += digits << position;
			} else {
				if (pair == 0) {
					digits = 0;
				} else if (pair == 1) {
					digits = 3;
				} else if (pair == 2) {
					digits = 1;
				} else {
					digits = 2;
				}
				res += digits << position;
			}

			a >>= 2;
			position += 2;
		}

		return res;
	}
}