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 collections import deque
from heapq import heappush, heappop

if __name__ == '__main__':
    t = int(input())
    for _ in range(t):
        n = int(input())
        i = 1
        heap = [(n - 1, (1, n))]
        result = [0] * n
        while heap:
            size, (curr_r, curr_l) = heappop(heap)
            idx = curr_l + curr_r
            summed = curr_l - curr_r + 1
            if summed % 2:
                idx = (curr_l + curr_r)//2
                result[idx - 1] = i
                i += 1
                if curr_r <= idx - 1:
                    heappush(heap, (-(idx - 1 - curr_r), (curr_r, idx - 1)))
                if curr_l >= idx + 1:
                    heappush(heap, (-(curr_l - idx - 1), (idx+1, curr_l)))
            else:
                idx = (curr_l + curr_r - 1)//2
                result[idx - 1] = i
                i += 1
                if curr_l >= idx + 1:
                    heappush(heap, (-(curr_l - idx - 1), (idx+1, curr_l)))
                if curr_r <= idx - 1:
                    heappush(heap, (-(idx - 1 - curr_r), (curr_r, idx - 1)))

        print(" ".join(map(str, result)))