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,stdout
from collections import deque
from heapq import heappush,heappop
Pi = lambda x: stdout.write(str(x) + '\n')
Ps = lambda x: stdout.write(str(x))
S = lambda x: x*(x+1) // 2
I = lambda x: 1+(2*x)
R = lambda:stdin.readline()
Ri = lambda x:map(int,x.split())
Rs = lambda x:map(str,x.split())
Rf = lambda x:map(float,x.split())
MaxN = int(1e5) + 5

def f(L,R,A):
	B = []
	heappush(B,(0,L,R))	
	cont = 0
	while len(B):
		x = heappop(B)
		L,R = x[1],x[2]
		cont += 1
		pos = (R + L) // 2
		A[pos] = cont		
		if L<pos:
			heappush(B,(-pos+L,L, pos - 1))
		if pos < R:
			heappush(B,(-R+pos,pos + 1, R))		
		
def main():
	A = [0]*(2*MaxN)
	T = int(R())
	for casos in range(T):
		n = int(R())		
		f(1,n,A)
		for i in range(1,n+1):
			Ps(str(A[i]) + ' ')
		Ps('\n')
		# Pi(cont)
if __name__=='__main__':
    main()