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 Main
{
	private void solve()throws IOException
	{
		int n=nextInt();
		int a[]=new int[n+1];
		PriorityQueue<Pair> pq=new PriorityQueue<>(11,new Comparator<Pair>(){
			public int compare(Pair a,Pair b){
				int sizea=a.second-a.first+1;
				int sizeb=b.second-b.first+1;
				return sizea==sizeb?a.first-b.first:sizeb-sizea;
			}
		});
		int c=1;
		pq.add(new Pair(1,n));
		while(!pq.isEmpty())
		{
			Pair p=pq.remove();
			int low=p.first,high=p.second,mid=(low+high)/2;
			if(low>high)
				continue;
			a[mid]=c++;
			pq.add(new Pair(low,mid-1));
			pq.add(new Pair(mid+1,high));
		}
		for(int i=1;i<=n;i++)
			out.print(a[i]+" ");
		out.println();
	}
	class Pair{
	    int first,second;
	    Pair(int a,int b){
	        first=a;
	        second=b;
	    }
	}

	 
	///////////////////////////////////////////////////////////

	public void run()throws IOException
	{
		br=new BufferedReader(new InputStreamReader(System.in));
		st=null;
		out=new PrintWriter(System.out);

		int t=nextInt();
		while(t-->0)
		solve();
		
		br.close();
		out.close();
	}
	public static void main(String args[])throws IOException{
		new Main().run();
	}
	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	String nextToken()throws IOException{
		while(st==null || !st.hasMoreTokens())
		st=new StringTokenizer(br.readLine());
		return st.nextToken();
	}
	String nextLine()throws IOException{
		return br.readLine();
	}
	int nextInt()throws IOException{
		return Integer.parseInt(nextToken());
	}
	long nextLong()throws IOException{
		return Long.parseLong(nextToken());
	}
	double nextDouble()throws IOException{
		return Double.parseDouble(nextToken());
	}
}