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 PYTHON):

Solution:
import heapq
for _ in range(input()):
    n=input()
    l=[0 for i in range(n+1)]
    s=[[-n,1]]
    heapq.heapify(s)
    rate=0
    for i in range(1,n+1):
        t=s[0]
        t.append(t[1]+abs(t[0])-1)
        if t[0]==-1:
            rate=i
            break
            
        g=abs(t[2])-abs(t[1])+1
        if g%2==0:
            pos=(abs(t[2])+abs(t[1])-1)/2
        else:
            pos=(abs(t[2])+abs(t[1]))/2
        l[pos]=i
        heapq.heappop(s)
        if g!=1:
            if pos!=abs(t[1]):
                heapq.heappush(s,[-((pos-1)-abs(t[1])+1),t[1]])
            if pos!=abs(t[2]):
                heapq.heappush(s,[-(abs(t[2])-(pos+1)+1),(pos+1)])
    for i in range(1,n+1):
        if l[i]!=0:
            print l[i],
        else:
            print rate,
            rate+=1
    print