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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int N = 100005;
const int first[3][4] = {{0, 1, 2, 3}, {0, 2, 3, 1}, {0, 3, 1, 2}};
ll n, m;
void solve() {
  scanf("%lld", &n);
  n--;
  m = n % 3;
  n /= 3;
  ll C = 0, D = 1;
  while (C + D <= n) {
    C += D;
    D *= 4;
  }
  n -= C;
  ll A = 0;
  for (ll i = 1; i < D; i *= 4) {
    A += first[m][(n / i) % 4] * i;
  }
  A += D * (m + 1);
  printf("%lld\n", A);
}
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) solve();
}
