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 stdout,stdin
from heapq import *
import io
import os
#input=stdin.readline
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
class Data:
    def __init__(self,val):
        self.val=val

    def __lt__(self,other):
        return (self.val[1]-self.val[0]+1>other.val[1]-other.val[0]+1) or(self.val[1]-self.val[0]+1==other.val[1]-other.val[0]+1 and self.val[0]<other.val[0])
if __name__=="__main__":
    for t in range(int(input())):
        n=int(input())
        a=[0 for i in range(n)]
        q=[Data([1,n])]
        heapify(q)
        i=1
        while q:
            x,y=heappop(q).val
            if (y-x+1)%2==1:
                a[(y+x)//2-1]=i
                if ((y+x)//2-1)>=x:
                    heappush(q,Data([x,(y+x)//2-1]))
                if (y + x) // 2 +1<=y:
                    heappush(q, Data([(y + x) // 2 +1,y]))
            else:
                a[(y+x-1)//2-1]=i
                if ((y + x-1) // 2 - 1)>=x:
                    heappush(q, Data([x, (y + x-1) // 2 - 1]))
                if ((y + x-1) // 2 + 1)<=y:
                    heappush(q, Data([(y + x-1) // 2 + 1, y]))
            i+=1
        print(*a)