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:
# cook your dish here
from sys import stdin
import heapq

t =  int(stdin.readline())

for _ in range(t):
    N =  int(stdin.readline())
    pattern = [0 for _ in range(N+1)]
    pattern_heap = [-N]
    pattern_hashmap = {N:[(1, N)]}

    for i in range(1, N+1) :
        max_zeros = pattern_heap[0]*(-1)
        l, r = heapq.heappop(pattern_hashmap[max_zeros])
        mid = (l+r)//2
        pattern[mid] = i
        if l < mid:
            if mid-l in pattern_hashmap:
                heapq.heappush(pattern_hashmap[mid-l], (l, mid-1))
            else:
                pattern_hashmap[mid-l] = [(l, mid-1)]
                heapq.heappush(pattern_heap, -(mid-l))
        if mid < r :
            if r-mid in pattern_hashmap:
                heapq.heappush(pattern_hashmap[r-mid], (mid+1, r))
            else:
                pattern_hashmap[r-mid] = [(mid+1, r)]
                heapq.heappush(pattern_heap, -(r-mid))


        if len(pattern_hashmap[max_zeros])==0:
            heapq.heappop(pattern_heap)
            del pattern_hashmap[max_zeros]

    for var in pattern[1:]:
        print(var, end=" ")
    print()