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
import heapq
for nt in range(int(input())):
	n=int(input())
	if n==1:
		print (1)
		continue
	if n==2:
		print (1,2)
		continue
	if True:
		ans=[-1]*n
		q2=[]
		ans[(n+1)//2-1]=1
		q2.append([-(n-((n+1)//2+1)+1),(n+1)//2+1,n])
		q2.append([-((n+1)//2)+1,1,(n+1)//2-1])
		heapq.heapify(q2)
		flag=1
		k=2
		while len(q2)!=0:
			group = heapq.heappop(q2)
			group[0] = -1*group[0]
			if group[0]==1:
				ans[group[1]-1]=k
			elif group[0]==2:
				ans[group[1]-1]=k
				heapq.heappush(q2,[-1,group[1]+1,group[2]])
			elif group[0]%2:
				ans[group[1]+group[0]//2-1]=k
				heapq.heappush(q2,([-(group[0]//2),group[1],group[1]+group[0]//2-1]))
				heapq.heappush(q2,([-(group[0]//2),group[1]+group[0]//2+1,group[2]]))
			else:
				ans[group[1]+group[0]//2-2]=k
				heapq.heappush(q2,([-(group[0]//2)+1,group[1],group[1]+group[0]//2-2]))
				heapq.heappush(q2,([-(group[0]//2),group[1]+group[0]//2,group[2]]))
			k+=1
		for i in range(n):
			if ans[i]==-1:
				ans[i]=k
				k+=1
		print (*ans)