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:
def bubble_sort(c):
    n = len(c)
    for i in range(n):
        already_sorted = True
        for j in range(n - i - 1):
            if c[j][0] < c[j+1][0]:#it is according to the no of zeros
                c[j][0], c[j+1][0] = c[j+1][0], c[j][0]
                already_sorted = False
            if(c[j][0]==c[j+1][0]):
            	if(c[j][1]>c[j+1][1]):#this is for frsts come find
            		c[j][1], c[j+1][1] = c[j+1][1], c[j][1]
        if already_sorted:
            break
    return c

def fun(l,r):
	if(l>r):
		return 
	m=(l+r)//2
	c.append([l-r,m])
	# c[m]=(r-l,l)
	fun(l,m-1)
	fun(m+1,r)
t=int(input())
while t:
	t-=1
	n=int(input())
	a=[0 for i in range(n)]
	c=[]
	fun(0,n-1)
	# c=bubble_sort(c)
	c.sort()
	# for i in range(n):
		# print(c[i][0],c[i][1],end=",")
	# print()
	for i in range(1,n+1):
		idi=c[i-1][1]
		a[idi]=i
	print(*a)