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.io.*;
import java.util.*;

public class D
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int t=Integer.parseInt(bu.readLine());
        while(t-->0)
        {
            int n=Integer.parseInt(bu.readLine()),i,a[]=new int[n+1];
            PriorityQueue<Node> pq=new PriorityQueue<>(new Comparator<Node>() {
                public int compare(Node o1,Node o2) {
                    if(o1.r-o1.l>o2.r-o2.l) return -1;
                    else if(o1.r-o1.l<o2.r-o2.l) return 1;
                    else
                    {
                        if(o1.l>o2.l) return 1;
                        else return -1;
                    }
                }});
            pq.add(new Node(1,n));
            for(i=1;i<=n;i++)
            {
                Node no=pq.poll();
                int m=0;
                if((no.r-no.l)%2==0)
                {
                    m=(no.r+no.l)/2;
                    a[m]=i;
                }
                else
                {
                    m=(no.r+no.l-1)/2;
                    a[m]=i;
                }
                if(no.l<m) pq.add(new Node(no.l,m-1));
                if(no.r>m) pq.add(new Node(m+1,no.r));
            }
            for(i=1;i<=n;i++)
                sb.append(a[i]+" ");
            sb.append("\n");
        }
        System.out.print(sb);
    }

    static class Node
    {
        int l,r;
        Node(int a,int b)
        {
            l=a; r=b;
        }
    }
}