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 sys
import heapq
from heapq import heappush, heappop

def main():
	q = []
	for line in sys.stdin.readlines():
		q.append(int(line.rstrip()))


	for i in range(1, len(q)):
		res = [0] * q[i]
		r = q[i]
		l = 1
		d = r - l + 1
		to_put = 1
		h = []
		heappush(h, (1, (l, r)) ) # doesnt matter what val is b/c its going to be popped
		while(len(h) != 0):
			# print(h)
			curr = heappop(h)
			pair = curr[1] 
			l = pair[0]
			r = pair[1]
			# print(pair)
			if (r - l + 1) % 2 == 0:
				middle = (l+r - 1) // 2
				res[ middle - 1 ] = str(to_put)
			else:
				middle = (l+r) // 2
				res [ middle - 1 ] = str(to_put)
			to_put +=1
			# append the splits
			heappush(h, (-1*(middle - l), (l, middle-1))) # push left
			heappush(h, (-1 * (r - middle), (middle+1, r))) # push right 
			if to_put == q[i]+1:
				break
		print(" ".join(res))
		

if __name__ == '__main__':
	main()