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, exit
from heapq import heappush, heappop
input = stdin.readline

def i(): return input()
def ii(): return int(input())
def iis(): return map(int, input().split())
def liis():	return list(map(int, input().split()))
def print_array(a): print(" ".join(map(str, a)))

def binary(a, l, r, i):
	if (r-l+1)%2 == 1:
		return (l+r)//2
	return (l+r-1)//2

def solve(a, l, r, i):
	ans = a
	tupla = (-len(a), 0, len(a)-1, ans)
	a = []
	heappush(a, tupla)
	#print(a)
	while len(a):
		size, ini, fim, cur = heappop(a)
		split = binary(cur, ini, fim, i)
		if ans[split] != 0: continue
		ans[split] = i
		#teste = input()
		left = ans[ini:split]
		right = ans[split+1:fim+1]
		#print(i, "cur, split, left, right: ", cur, split, left, right)
		if len(right) > len(left):
			if len(right) > 0:
				heappush(a, (-len(right), split+1, fim, right))
			if len(left) > 0:
				heappush(a, (-len(left), ini, split-1, left))
		else:
			if len(left) > 0:
				heappush(a, (-len(left), ini, split-1, left))
			if len(right) > 0:
				heappush(a, (-len(right), split+1, fim, right))
		#print(a)
		#a = deque(sorted(a))
		i += 1
	return ans
	


t = ii()
for _ in range(t):
	n = ii()
	a = [0]*n	
	a = solve(a, 0, n-1, 1)
	#print("ANSWER")
	print_array(a)