B. Roof Construction

time limit per test

1 second

memory limit per test

256 megabytes

It has finally been decided to build a roof over the football field in School 179. Its construction will require placing n consecutive vertical pillars. Furthermore, the headmaster wants the heights of all the pillars to form a permutation p of integers from 0 to n−1, where pi is the height of the i-th pillar from the left (1≤i≤n).

As the chief, you know that the cost of construction of consecutive pillars is equal to the maximum value of the bitwise XOR of heights of all pairs of adjacent pillars. In other words, the cost of construction is equal to max1≤i≤n−1pi⊕pi+1, where ⊕ denotes the bitwise XOR operation.

Find any sequence of pillar heights p of length n with the smallest construction cost.

In this problem, a permutation is an array consisting of n distinct integers from 0 to n−1 in arbitrary order. For example, [2,3,1,0,4] is a permutation, but [1,0,1] is not a permutation (1 appears twice in the array) and [1,0,3] is also not a permutation (n=3, but 3 is in the array).

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤104). Description of the test cases follows.

The only line for each test case contains a single integer n (2≤n≤2⋅105) --- the number of pillars for the construction of the roof.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105.

Output

For each test case print n integers p1, p2, ..., pn --- the sequence of pillar heights with the smallest construction cost.

If there are multiple answers, print any of them.

Example

Input

Copy

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Output

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0 1
2 0 1
3 2 1 0 4
4 6 3 2 0 8 9 1 7 5

Note

For n=2 there are 2 sequences of pillar heights:

$0,1\] --- cost of construction is 0⊕1=1.$

For n=3 there are 6 sequences of pillar heights:

$0,1,2\] --- cost of construction is max(0⊕1,1⊕2)=max(1,3)=3.$
$1,0,2\] --- cost of construction is max(1⊕0,0⊕2)=max(1,2)=2.$
$2,0,1\] --- cost of construction is max(2⊕0,0⊕1)=max(2,1)=2.$

解题说明：此题是一道数学题，为了保证值最小，找规律能发现要让相邻两个数的高位尽量相等。

cpp 复制代码

#include <stdio.h>

int main() 
{
    int tt;
    scanf("%d", &tt);
    while (tt--) 
    {
        int n;
        scanf("%d", &n);
        int k = 1;
        while (k < n) 
        {
            k *= 2;
        }
        k /= 2;
        for (int i = 1; i < n; i++) 
        {
            if (i == k) 
            {
                printf("0 %d ", k);
            }
            else
            {
                printf("%d ", i);
            }
        }
        printf("\n");
    }
    return 0;
}