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:
#qn given an arr of size n for every  i position find a min j such that j>i and arr[j]>arr[i]
#it prints the position of the nearest .
import sys
input = sys.stdin.readline
 #qn given an arr of size n for every  i position find a min j such that j>i and arr[j]>arr[i]
#it prints the position of the nearest .
# import sys
import heapq 
import copy
import math
#heapq.heapify(li) 
#
#heapq.heappush(li,4) 
#
#heapq.heappop(li)
#
# &	Bitwise AND Operator	10 & 7 = 2
# |	Bitwise OR Operator	10 | 7 = 15
# ^	Bitwise XOR Operator	10 ^ 7 = 13

# <<	Bitwise Left Shift operator	10<<2 = 40
# >>	Bitwise Right Shift Operator
############ ---- Input Functions ---- #######Start#####
def inp():
	return(int(input()))
def inlt(): 
	return(list(map(int,input().split())))
def insr():
	s = input()
	return(list(s[:len(s) - 1]))
def invr():
	return(map(int,input().split()))
 ############ ---- Input Functions ---- #######End
 # #####   


class Node:
    def _init_(self,val):
        self.data = val
        self.left = None
        self.right = None

def bin_search(arr,l,r,val):#strickly greater
	if arr[r] <= val:
		return r+1	
	if r-l < 2:
		if arr[l]>val:
			return l
		else:
			return r
	mid = int((l+r)/2)
	if arr[mid] <= val:
		return bin_search(arr,mid,r,val)
	else:
		return bin_search(arr,l,mid,val)

def pr_list(a):
	print(*a, sep=" ")
def find_permut(hashm,l):
	ans = 1
	total = math.factorial(l)
	for i in hashm:
		if i!=0:
			ans = ans*math.factorial(i)
	return int(total/ans)
def main():
	tests =  inp()
	mod = 1000000007
	limit = 10**18
	# print(limit)

	for test in range(tests):
		# ans = 0 
		n = inp()
		ans = [0 for i in range(n)]
		heap = [[-1*n,0,n-1]]
		heapq.heapify(heap)

		k = 1
		while(heap != []):
			[val,i,j] = heapq.heappop(heap)
			mid = i + int((j-i)/2)
			ans[mid] = k
			k = k+1
			if (j-i +1)%2 != 0 and j !=i:
				heapq.heappush(heap,[-1*(mid - i),i,mid-1])
				heapq.heappush(heap,[-1*(j - mid),mid+1,j])
			elif (j-i + 1)%2 == 0 and j-i!=1:
				heapq.heappush(heap,[-1*(mid - i),i,mid-1])
				heapq.heappush(heap,[-1*(j - mid),mid+1,j])
			elif j-i==1:
				heapq.heappush(heap,[-1*(j - mid),mid+1,j])
		pr_list(ans)

		# print(ans)

if __name__== "__main__":
  main()