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 heapq 

t = int(input())

class ListNode:
    def __init__(self, l, r):
        self.l = l
        self.r = r

    def __lt__(self, other):
        x1, y1 = self.l, self.r
        x2, y2 = other.l, other.r

        dia1, dia2 = y1 - x1, y2 - x2

        if dia1 > dia2:
            return True
        elif dia1 < dia2:
            return False
        else:
            return x1 < x2


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

    queue = []
    heapq.heappush(queue, ListNode(0, n - 1))
    
    # setattr(ListNode, "__lt__", lambda self, other: self.val <= other.val)

    idx = 1
    while queue:
        node = heapq.heappop(queue)
        l, r = node.l, node.r

        #print('checking', l, r)

        k = 0
        if (r - l + 1) % 2 == 1:
            a[(l + r) // 2] = idx
            k = (l + r) // 2
        else:
            a[(l + r - 1) // 2] = idx
            k = (l + r - 1) // 2
        
        idx += 1

        if l == r:
            continue

        l1, r1 = l, k - 1
        l2, r2 = k + 1, r

        if r2 - l2 >= 0:
            heapq.heappush(queue, ListNode(l2, r2))
        
        if r1 - l1 >= 0:
            heapq.heappush(queue, ListNode(l1, r1))

        # if r2 - l2 > r1 - l1:
        #     if r2 - l2 >= 0:
        #         heapq.heappush((l2, r2))
        #         #queue.append((l2, r2))
        #     if r1 - l1 >= 0:
        #         queue.append((l1, r1))
        # else:
        #     if r1 - l1 >= 0:
        #         queue.append((l1, r1))
        #     if r2 - l2 >= 0:
        #         queue.append((l2, r2))
        #         heapq.heappush((l2, r2))

        

        #print(' '.join(list(map(str, queue))))

    print(' '.join(list(map(str, a))))

for _ in range(t):
    n = int(input())

    solve(n)