Problem: You are given an array a of length n consisting of zeros. You perform n actions with this array: during the i-th action, the following sequence of operations appears:

  1. Choose the maximum by length subarray (continuous subsegment) consisting only of zeros, among all such segments choose the leftmost one; 
  2. Let this segment be [l; r]. If r-l+1 is odd (not divisible by 2) then assign (set) a[(l+r)/(2)] := i (where i is the number of the current action), otherwise (if r-l+1 is even) assign (set) a[(l+r-1)/(2)] := i. 



Consider the array a of length 5 (initially a=[0, 0, 0, 0, 0]). Then it changes as follows:

  1. Firstly, we choose the segment [1; 5] and assign a[3] := 1, so a becomes [0, 0, 1, 0, 0]; 
  2. then we choose the segment [1; 2] and assign a[1] := 2, so a becomes [2, 0, 1, 0, 0]; 
  3. then we choose the segment [4; 5] and assign a[4] := 3, so a becomes [2, 0, 1, 3, 0]; 
  4. then we choose the segment [2; 2] and assign a[2] := 4, so a becomes [2, 4, 1, 3, 0]; 
  5. and at last we choose the segment [5; 5] and assign a[5] := 5, so a becomes [2, 4, 1, 3, 5]. 



Your task is to find the array a of length n after performing all n actions. Note that the answer exists and unique.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the array a of length n after performing n actions described in the problem statement. Note that the answer exists and unique.

Example

Input


6
1
2
3
4
5
6


Output


1 
1 2 
2 1 3 
3 1 2 4 
2 4 1 3 5 
3 4 1 5 2 6 

Solution (in PYTHON3):

Solution:
import functools
import heapq


class Section:
    def __init__(self, left, right):
        self.left = left
        self.right = right
        self.size = right - left
        if (right - left + 1) % 2 != 0:
            self.aim = (left + right) // 2
        else:
            self.aim = (left + right - 1) // 2

    @functools.total_ordering
    def __lt__(self, other):
        if self.size > other.size:
            return True
        if self.size < other.size:
            return False
        return self.left < other.left

    @functools.total_ordering
    def __eq__(self, other):
        return self.size == other.size and self.left == self.right


def solution(n):
    h = []
    res = [0 for i in range(n)]
    cnt = 1
    sec = Section(0, n - 1)
    heapq.heappush(h, sec) # добавили в кучу первый отрезок, потому что все нули
    while len(h) > 0:
        current = heapq.heappop(h) # вытащили самый длинный левый отрезок с нулями
        res[current.aim] = cnt # присвоили в ответ номер операции
        cnt += 1
        if current.left != current.aim:
            new_sec1 = Section(current.left, current.aim - 1)
            heapq.heappush(h, new_sec1)
        if current.right != current.aim:
            new_sec2 = Section(current.aim + 1, current.right)
            heapq.heappush(h, new_sec2)
    print(' '.join(map(str, res)))


inp = int(input())
for _ in range(inp):
    n = int(input())
    solution(n)