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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops, no-stack-protector")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const long long maxn = 2e5 + 30;
const long long mod = 1e9 + 7;
const long long base = 1e9;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (fopen("t.inp", "r")) {
    freopen("test.inp", "r", stdin);
    freopen("test.out", "w", stdout);
  }
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long nw = 1;
    while (n >= nw) nw *= 4;
    nw /= 4;
    long long cl = n - (nw - 1);
    long long pos = (cl - 1) / 3 + 1;
    long long a = pos + nw - 1;
    long long b = 0;
    for (int i = 0; i < 60; i += 2) {
      long long nw = 0;
      if (a & (1ll << i)) nw++;
      if (a & (1ll << (i + 1))) nw += 2;
      if (nw == 1)
        b += (1ll << (i + 1));
      else if (nw == 2)
        b += (1ll << (i)), b += (1ll << (i + 1));
      else if (nw == 3)
        b += (1ll << (i));
    }
    long long c = a ^ b;
    if (n % 3 == 1)
      cout << a << "\n";
    else if (n % 3 == 2)
      cout << b << "\n";
    else
      cout << c << "\n";
  }
}
