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:
from heapq import heapreplace, heapify, heappop
from bisect import insort


def merge(iterables):
    "Merge sorted iterables."
    entries = []          # Heap of [front value, id, iterator].
    for id, it in enumerate(map(iter, iterables)):
        try:
            entries.append([next(it), id, it])
        except StopIteration:
            pass
    heapify(entries)
    while entries:
        value, _, it = entry = entries[0]
        yield value
        try:
            entry[0] = next(it)
            heapreplace(entries, entry)
        except StopIteration:
            heappop(entries)


def combine(a, b):
    a, b = iter(a), iter(b)

    x, y = next(a, none), next(b, none)
    while x or y:
        if x is none:
            yield y
            y = next(b, none)
        elif y is none:
            yield x
            x = next(a, none)
        else:
            if x < y:
                yield x
                x = next(a, none)
            else:
                yield y
                y = next(b, none)


def construct(n):
    a = [0] * n

    queue = [-1, n]
    index = []
    g = iter(range(1, n + 1))
    i = 0
    while len(queue) - 2 != n:
        level, prev = [], queue[0]
        for y in queue:
            x = prev
            if x + 1 < y:
                i = (x + y) // 2
                index.append((x - y, i))
                level.append(i)
            prev = y

        queue = list(merge([iter(queue), iter(level)]))

    for _, i in sorted(index):
        a[i] = next(g)
    return a

def solve():
    n = int(input())
    print(*construct(n))


for _ in range(int(input())):
    solve()