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

class HeapNode: 
    def __init__(self, i, j): 
        self.i = i
        self.j = j 
        self.count = j-i+1 

def test(a,b):  
    if a.count == b.count: 
        return a.i < b.i 
    else: 
        return a.count > b.count 

setattr (HeapNode, "__lt__", lambda self, other: test(self, other))  

t = int(input()) 
for _ in range (t): 
    
    n = int(input()) 
    out = [0]*n  
    i = 1 

    heap = list() 
    heapq.heappush(heap, HeapNode (0, n-1)) 

    while heap:  

        x = heapq.heappop (heap)  
        if x.count%2 == 0:  
            pivot = (x.i + x.j -1)//2 
        else: 
            pivot = (x.i + x.j)//2 

        out[pivot] = i 
        i += 1 

        l1,r1 = x.i,pivot-1 
        l2,r2 = pivot+1,x.j 

        if r1-l1+1 > 0: 
            heapq.heappush (heap,HeapNode (l1,r1)) 
        if r2-l2+1 > 0:
            heapq.heappush (heap,HeapNode (l2,r2))    

    for o in out: 
        print (o, end=" ") 
    print()