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 JAVA):

Solution:
import java.util.*;
public class check1 {

    static class pair
    {
        int s;
        int e;
        pair(int s,int e)
        {
            this.s=s;
            this.e=e;
        }
    }
    public static void find1(int[] a,int s,int e)
    {
        PriorityQueue<pair> q=new PriorityQueue<>(new Comparator<pair>() {
            @Override
            public int compare(pair o1, pair o2) {
                int a=o1.e-o1.s;
                int b=o2.e-o2.s;
                if(a!=b) return b-a;
                else return o1.e-o2.e;
            }
        });
        q.add(new pair(s,e));

        int mid=s+(e-s)/2;
        int count=1;
        while(!q.isEmpty())
        {
            pair t1=q.poll();
            s=t1.s;
            e=t1.e;

            if(s>e) continue;

            mid=s+(e-s)/2;
            a[mid]=count++;
            q.add(new pair(s,mid-1));
            q.add(new pair(mid+1,e));
        }

    }
    public static void main(String[] args) {

        Scanner sc=new Scanner(System.in);
        int tt=sc.nextInt();
        while(tt-- >0)
        {
            int n=sc.nextInt();
            int a[]=new int[n];
            int s=0;
            int e=n-1;
            find1(a,s,e);
            for(int i=0;i<n;i++){System.out.print(a[i]+" ");}
            System.out.println();
        }

    }
//41664916690999888
}