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 sys import stdin
from math import sqrt
input=stdin.readline
for _ in range(int(input())):
	n=int(input())
	d={n:[(n+1)//2]}
	v=[0]*(n+2)
	stck=[(n+1)//2]
	v[(n+1)//2]=v[0]=v[-1]=1
	while stck:
		p=stck.pop()
		l=r=p+1
		while v[r]==0:
			r+=1
		r-=1
		if v[l]==0:
			m=(l+r)//2
			stck+=m,
			v[m]=1
			if r-l+1 not in d:d[r-l+1]=[m]
			else:d[r-l+1]+=m,
		l=r=p-1
		while v[l]==0:
			l-=1
		l+=1
		if v[r]==0:
			m=(l+r)//2
			stck+=m,
			v[m]=1
			if r-l+1 not in d:d[r-l+1]=[m]
			else:d[r-l+1]+=m,
	v=[0]*n;p=1
	for i in sorted(d,reverse=True):
		for j in sorted(d[i]):
			v[j-1]=p
			p+=1
	print(*v)