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:


# 1 
# 1 2 
# 2 1 3 
# 3 1 2 4 
# 2 4 1 3 5 

# 3 4 1 5 2 6 

# 4	2 5	1 6 3 7 




from sys import stdin ,stdout
from pprint import pprint 
from heapq import heappush ,heappop


for _ in range(int(stdin.readline())):
	n=int(input())
	ans=[-1]*(n+1)

	cnt=1
	heap=[(-(n-1+1),1,n)] #(l-r+1,l,r)
	while(cnt<=n):
		# print(heap)
		node=heappop(heap)
		l=node[1]
		r=node[2]
		if (r>=l):
			pos=l+(r-l)//2
			ans[pos]=cnt
			cnt+=1
			heappush(heap,(-(pos-1-l+1),l,pos-1))
			heappush(heap,(-(r-(pos+1)+1),pos+1,r))

	for i in ans[1:]:
		stdout.write(str(i)+' ')
	stdout.write('\n')