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,stdout
import heapq
for testcases in range(int(stdin.readline())):
    n=int(stdin.readline())
    if n==1 :
        print(1)
    elif n==2:
        print(1,2)
    else:
        k=1
        arr=[0 for i in range(n)]
        store= [(-n, ( 0,n-1))]
        chance = 1
        heapq.heapify(store)
        while(len(store)>0):
            # store=(sorted(store,reverse=False))
            # print('=========================================')
            # print(store)
            value=heapq.heappop(store)
            # print(value)
            # print(store)
            # store=set(store)
            left = value[1][0]
            right=value[1][1]
            if left > right :
                continue
            if left==right :
                arr[left]=k
                k+=1
            else :
                if ( right - left +1 ) % 2==0:
                    # print('first')
                    mid = (left + right -1) //2
                    arr[mid]=k
                    k+=1
                    heapq.heappush(store,(-(right -mid),(mid+1 ,right)))
                    heapq.heappush(store,(-(mid - left ) ,(left , mid-1 )))
                else:
                    mid = (left + right) //2
                    arr[mid]=k
                    # print('second')
                    k+=1
                    heapq.heappush(store,(-(right - mid ) , ( mid +1 , right)))
                    heapq.heappush(store,(-(mid - left ) , ( left , mid-1)))

            # print(arr)

        for i in arr:
            stdout.write(str(i) + ' ')
        stdout.write('\n')