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:
t = int(input())
for i in range(t):
    num = int(input())
    ans = [0]*num
    l = 0
    count = 1
    data = {num : [l]}
    lisp = [num]
    flag = 0
    for i in lisp:
        subseqs = sorted(data[i])
        n = i
        for j in subseqs:
            l = j
            r = l+n-1
            if n%2==0:  #For even addition
                
                pos = l+(n//2)-1  #Position to add number
                if (n//2) not in lisp:
                    lisp.append(n//2) 
                if ((n//2)-1) not in lisp:    
                    lisp.append((n//2)-1)
                
                if (n//2) in data:  #Adding the larger one
                    data[n//2].append(pos+1)
                else:
                    data[n//2] = [pos+1]
                
                if ((n//2)-1) in data: #Adding the smaller division
                    data[(n//2)-1].append(l)
                else:
                    data[(n//2)-1] = [l]
                    
                ans[pos] = count
                count += 1
                
            else: #For odd addition
                pos = l + (n//2)
                if (n//2) not in lisp:
                    lisp.append(n//2)
                if (n//2) in data:  
                    data[n//2].append(l)
                    data[n//2].append(pos+1)
                else:
                    data[n//2] = [l , (pos+1)]
                
                ans[pos] = count
                count += 1
            
            
            if count == num+1:
                flag = 1
                break
        if flag == 1:
            break
    print(*ans)