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, ... 
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000;
const int ff[4] = {0, 2, 3, 1};
int main() {
  int cas;
  scanf("%d", &cas);
  while (cas--) {
    long long n;
    scanf("%lld", &n);
    long long g = (n - 1) / 3 + 1;
    int id = (n - 1) % 3 + 1;
    int B = 0;
    long long N = 1;
    while (g > N) {
      g -= N;
      N *= 4;
      B++;
    }
    g--;
    long long x = N + g;
    long long y = 1ll << (B * 2 + 1);
    for (int i = (0); i <= (B - 1); i++) {
      long long t = (x >> (i * 2)) & 3;
      long long tt = ff[t];
      y += tt << (i * 2);
    }
    long long z = x ^ y;
    if (id == 1) printf("%lld\n", x);
    if (id == 2) printf("%lld\n", y);
    if (id == 3) printf("%lld\n", z);
  }
  return 0;
}
