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:
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import heapq
t=int(input())
for i in range(t):
    n=int(input())
    arr=[[(-1*n),0]]
    ans=[0]*n
    j=1
    s=0
    while s<n and arr[0][0]!=0:
        lis=heapq.heappop(arr)
        temp=abs(lis[0])
        l=lis[1]
        if temp%2==0:
            ans[l+(temp//2)-1]=j
            if temp!=2:
                heapq.heappush(arr,[(-1)*((temp//2)-1),l])
                heapq.heappush(arr,[(-1)*(temp//2),l+(temp//2)])
            else:
                heapq.heappush(arr,[0,l])
                heapq.heappush(arr,[(-1)*(temp//2),l+(temp//2)])
                s+=1
        else:
            ans[l+(temp//2)]=j
            if temp!=1:
                heapq.heappush(arr,[(-1)*(temp//2),l])
                heapq.heappush(arr,[(-1)*(temp//2),l+(temp//2)+1])
            else:
                heapq.heappush(arr,[0,l])
                s+=1
        j+=1
    print(" ".join(str(x) for x in ans))