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, ... 
memo=[(0,0,0)]

def new_find_pair(n):
    stack=[]
    while n>=len(memo):
        N=n.bit_length()-1
        if N%2==0:
            i=N//2
            stack.append((n,1,i))
            n-=4**i
        else:
            i=N//2
            if n>>(2*i)&1==0:
                #type2
                stack.append((n,2,i))
                n-=2*4**i
            else:
                #type3
                stack.append((n,3,i))
                n-=3*4**i
    a,b,c=memo[n]
    for m,t,i in stack[::-1]:
        m-=t*4**i
        if t==1:
            if m==b:
                a,b,c=b,c,a
            elif m==c:
                a,b,c=c,a,b
        elif t==2:
            if m==a:
                a,b,c=c,a,b
            elif m==c:
                a,b,c=b,c,a
        else:
            if m==a:
                a,b,c=b,c,a
            elif m==b:
                a,b,c=c,a,b
        a+=4**i
        b+=2*4**i
        c+=3*4**i
    return (a,b,c)

for i in range(1,4**5):
    memo.append(new_find_pair(i))

def nth_pair(n):
    tmp=0
    for i in range(28):
        tmp+=4**i
        if tmp>=n:
            tmp-=4**i
            first=(n-tmp-1)+4**i
            return new_find_pair(first)

def nth_seq(n):
    q=(n-1)//3+1
    fst=(n-1)%3
    res=nth_pair(q)[fst]
    return res

import sys

input=sys.stdin.buffer.readline

#n=int(input())
#print(find_pair(n))
#print(nth_pair(n))
Q=int(input())
for _ in range(Q):
    n=int(input())
    #n=int(input())
    r=nth_seq(n)
    print(r)
    #print(nth_pair(q)[fst])
#print(time.time()-start)
