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 a[100010];
long long pow(long long x, long long y, long long k) {
  long long res = 1;
  while (y) {
    if (y & 1) res *= x;
    res = res % k;
    if (x > 1e9) break;
    x *= x;
    x = x % k;
    y = y >> 1;
  }
  return res;
}
long long c[200];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long n;
    scanf("%lld", &n);
    long long o = 0, res = 1, ans = 0;
    long long now = 0;
    while (((pow(4, o + 1, 1e18) - 1) / 3) * 3 < n) {
      o++;
    }
    o--;
    long long m;
    m = n - ((pow(4, o + 1, 1e18) - 1) / 3) * 3;
    long long p = (m - 1) / 3, pmod = m % 3;
    if (pmod == 0) pmod = 3;
    ans += (pmod << 2 * (o + 1));
    long long flag = p % 4;
    o = 0;
    while (p > 0) {
      c[o] = p % 4;
      o++;
      p = p / 4;
    }
    for (long long i = 0; i < o; i++) {
      if (c[i] != 0) {
        c[i] = (long long)(c[i] + pmod - 1) % 3;
        if (c[i] == 0) c[i] = 3;
      }
      ans += ((long long)c[i] << (long long)(2 * i));
    }
    printf("%lld\n", ans);
  }
}
