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;
long long ads[4][3];
pair<long long, pair<long long, long long> > ini_ans[256];
pair<long long, pair<long long, long long> > solve(long long x) {
  if (x < 256) return ini_ans[x];
  int r = x & 3;
  pair<long long, pair<long long, long long> > r1 = solve(x / 4);
  long long x1 = r1.first * 4 + ads[r][0];
  long long x2 = r1.second.first * 4 + ads[r][1];
  long long x3 = r1.second.second * 4 + ads[r][2];
  return pair<long long, pair<long long, long long> >(
      x1, pair<long long, long long>(x2, x3));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  set<int> s;
  vector<int> v;
  for (int i = 1; i < 150; i++) {
    if (s.find(i) != s.end()) continue;
    for (int j = i + 1; 1; j++) {
      if (s.find(j) != s.end()) continue;
      if (s.find(i ^ j) != s.end()) continue;
      s.insert(i);
      s.insert(j);
      s.insert(j ^ i);
      v.push_back(i);
      v.push_back(j);
      v.push_back(j ^ i);
      ini_ans[i] = pair<long long, pair<long long, long long> >(
          i, pair<long long, long long>(j, j ^ i));
      break;
    }
  }
  ads[1][0] = 1;
  ads[1][1] = 2;
  ads[1][2] = 3;
  ads[2][0] = 2;
  ads[2][1] = 3;
  ads[2][2] = 1;
  ads[3][0] = 3;
  ads[3][1] = 1;
  ads[3][2] = 2;
  int t;
  cin >> t;
  while (t--) {
    long long x;
    cin >> x;
    if (x < 256)
      cout << (v[x - 1]) << '\n';
    else {
      long long v = 256;
      while (v <= x) v *= 4;
      v /= 4;
      x -= v;
      int t = x % 3;
      long long vv = x / 3 + v;
      pair<long long, pair<long long, long long> > res = solve(vv);
      if (t == 0)
        cout << (res.first) << '\n';
      else if (t == 1)
        cout << (res.second.first) << '\n';
      else
        cout << (res.second.second) << '\n';
    }
  }
  return 0;
}
