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())
d = 0

def BFS(l, r):
    global d
    d = 1
    Q = []
    heapq.heappush(Q, (-(r - l), l, r))

    arr = [0] * r

    while len(Q) > 0:
        x, l, r = heapq.heappop(Q)
        m = None

        if (r - l + 1) % 2 == 1:
            m = (l + r) // 2

            if m - 1 >= l:
                heapq.heappush(Q, (-(m - 1 - l), l, m - 1))

            if r >= m + 1:
                heapq.heappush(Q, (-(r - m - 1), m + 1, r))

        else:
            m = (l + r - 1) // 2

            if r >= m + 1:
                heapq.heappush(Q, (-(r - m - 1), m + 1, r))

            if m - 1 >= l:
                heapq.heappush(Q, (-(m - 1 - l), l, m - 1))


        arr[m - 1] = d
        d += 1

        # print(l, r, m, ' --> ', arr)
    return arr


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

    print(*BFS(1, n))