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:
import math
t = int(input())


for ti in range(t):
	n = int(input())
	# len of a
	a = [0]*n

	loops = int(math.log(n,2))
	total = 0
	bounds = [(0,n-1)]
	singles = []
	for i in range(loops+1):
		newbounds = []
		for j in range(min(2**i,len(bounds))):
			if total == n:
				break
			change = int((bounds[j][0]+bounds[j][1])/2)
			a[change] = 2**i+j
			total += 1
			if change-1-bounds[j][0] > 0:
				newbounds.append((bounds[j][0],change-1))
			elif change-1-bounds[j][0] == 0:
				singles.append((bounds[j][0],change-1))
			if -change-1+bounds[j][1] > 0:
				newbounds.append((change+1,bounds[j][1]))
			elif -change-1+bounds[j][1] == 0:
				singles.append((change+1,bounds[j][1]))

		newbounds.sort(key=lambda x: (x[1]-x[0],-x[0]), reverse=True)
		bounds = newbounds

	singles.sort(key=lambda x: (x[1]-x[0],-x[0]), reverse=True)
	for ind,s in enumerate(singles):
		change = int((s[0]+s[1])/2)
		a[change] = n - len(singles) + ind + 1
			

	print(*a)


# try:

	# raise Exception
# except:
	# print("-1")
	


# thenos.sort(key=lambda x: x[2], reverse=True)

# int(math.log(max(numbers)+1,2))
# 2**3 (power)


# 5 -> 2 0 3 1 4
# 7 -> 3 1 5 0 2 4 6
# 9 -> 
# 10-> 4 1 7 2 5 8 0 3 6 9