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:
t=int(input())	
for i in range(t):
	n=int(input())
	d={n:{0:n-1}}
	l=[0]*n
	m=n
	l1=[n]
	k=0
	v=0
	while(len(d)>0):
		l2=[]
		if m==0:
			break
		l3=sorted(d[m])
		for v1 in l3:
			j=(v1,d[m][v1])
			k+=1
			if (j[1]-j[0]+1)%2==0:
				l[(j[0]+j[1]-1)//2]=k
				c=(j[0]+j[1]-1)//2
			else:
				l[(j[0]+j[1])//2]=k
				c=(j[0]+j[1])//2
			if j[1]-c>0:		
				if j[1]-c not in d:
					d[j[1]-c]={}
					l1.append(j[1]-c)
			if c-j[0]>0:
				if c-j[0] not in d:
					d[c-j[0]]={}
					l1.append(c-j[0])
			"""if j[1]-c>0 and c-j[0]>0 and j[1]-c==c-j[0]:
				x1=0
				for b in d[j[1]-c]:
					if b[0]<j[0]:
						x1=1
					break
				if x1==1:
					d[c-j[0]][j[0]]=c-1
					d[c-j[0]][c+1]=j[1]
					continue	"""		
			if j[1]-c>0:
				"""d1={j[1]-c:{}}
				x=0
				for b in d[j[1]-c]:
					if b[0]>c+1:
						x=1
					break					
				d1[j[1]-c][(c+1,j[1])]=1	
				if x==1:
					d1[j[1]-c].update(d[j[1]-c])
					d[j[1]-c]=d1[j[1]-c]
				else:"""
				d[j[1]-c][c+1]=j[1]
			if c-j[0]>0:
				"""d1={c-j[0]:{}}		
				x=0
				for b in d[c-j[0]]:
					if b[0]>j[0]:
						x=1
					break	
				d1[c-j[0]][(j[0],c-1)]=1
				if x==1:
					d1[c-j[0]].update(d[c-j[0]])
					d[c-j[0]]=d1[c-j[0]]
				else:"""
				d[c-j[0]][j[0]]=c-1
		v+=1
		if v<len(l1):
			m=l1[v]
		else:
			m=0
	print(*l)