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 heapq

t=int(input())
for _ in range(t):
    n=int(input())
    cand=[]
    q=[]
    heapq.heappush(q,(-n,1))
    while len(q)!=0:
        s,l=heapq.heappop(q)
        s*=-1
        if s==1:
            cand.append([s,l])
        elif s==2:
            cand.append([s,l])
            heapq.heappush(q,(-1,l+1))
        else:
            cand.append([s,l])
            if s%2==0:
                heapq.heappush(q,(-(s//2-1),l))
                heapq.heappush(q,(-(s//2),l+s//2))
            else:
                heapq.heappush(q,(-(s//2),l))
                heapq.heappush(q,(-(s//2),l+(s+1)//2))
    cand=sorted(cand,key=lambda x:x[1])
    cand=sorted(cand,reverse=True,key=lambda x:x[0])
    ans=[0]*n
    for i in range(n):
        pos=(cand[i][0]-1)//2+cand[i][1]
        ans[pos-1]=i+1
    print(*ans)