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 re
def find_all(arr, sub):
    a=[]
    for county in range(len(arr)-sub+1):
        if arr[county:county+sub]==[0]*sub:
            a.append(county)
    return a

def solve(arr,n):
    count=1
    while count<=n:
        l0=0
        current=0
        indexes=[]
        for i in range(n):
            if arr[i]==0:
                current+=1
                if i==n-1:
                    if current>l0:
                        l0=current
                        indexes=[i-l0+1]
                    elif current==l0:
                        indexes.append(i-l0+1)
                    else:
                        pass
            else:
                if current>l0:
                    l0=current
                    indexes=[i-l0]
                elif current==l0:
                    indexes.append(i-l0)
                else:
                    pass
                current=0  
        # indexes=list(find_all(arr,l0))
        for i in indexes:
            if count>n:
                break
            if i+l0%2==0:
                arr[(i+(i+l0-1)-1) //2]=count
            else:
                arr[ (i+(i+l0-1)) //2 ]=count
            count+=1
    return arr
        
        

for t in range(int(input())):
    n=int(input())
    arr=[0 for _ in range(n)]
    print(*solve(arr,n))