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 atexit
import io
import sys
from collections import deque 
from heapq import * 
 
_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__

 
 
def fun(a,b):
	if((b-a+1 )% 2 == 1) :
		m = (b+a) // 2  
	else :
		m = (b+a-1) // 2
	return m 
 
# print(pur(1,5))
tc = int(input()) 
pur = fun
for zzz in range(tc) :
	n = int(input()) 
	arr = []
	arr.append(0) 
	for i in range(n) :
		arr.append(0)  
	
	dq1 = deque() 
	dq2 = deque() 
 
	dq1.append((1,n)) 
 
	while(len(dq1) > 0):
		tem = dq1.popleft() 
		dq2.append((-(tem[1]-tem[0]),tem[0],tem[1]))
		a = tem[0] 
		b = tem[1] 
		
		if(a != b) :
			m = pur(a,b)
			if(a != m) :
				dq1.append((a,m-1))
			if(b != m) :
				dq1.append((m+1,b)) 
			
 
	# print(dq2)
	oww = [] 
 
	for ele in dq2 :
		heappush(oww,ele)
 
	i = 1
	while (len(oww) > 0):
		r = heappop(oww)
		d = pur(r[1],r[2])
		arr[d] = i 
		i += 1 
 
	yum = [str(i) for i in arr[1:]]
	print(" ".join(yum))