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.*;
import java.io.*;
public class Codeforces {
	private static class Pair implements Comparable<Pair>{
		int start;
		int end;
		
		Pair(int a, int b)
		{
			this.start=a;
			this.end=b;
		}
		
		public int compareTo(Pair other)
		{
			int l1=this.end-this.start+1;
			int l2=other.end-other.start+1;
			
			if(l1==l2) {
				return this.start-other.start;
			}
			return l2-l1;
		}
	}
	public static void main(String[]args) throws IOException
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

		    int tc=Integer.parseInt(br.readLine());
		    for(int t=0;t<tc;t++) {
		    	
		    	int n=Integer.parseInt(br.readLine());
		    	
		    	PriorityQueue<Pair>PQ=new PriorityQueue<>();
		    	
		    	PQ.add(new Pair(0,n-1));
		    	int count=0;
		    	int[]A=new int[n];
		    	
		    	while(count<n)
		    	{
		    		   Pair rem=PQ.remove();
		    		   int pos=0;
		    		   if((rem.end-rem.start+1)%2!=0)
		    		   pos=(rem.end+rem.start)/2;
		    		   
		    		   else
		    		   pos=(rem.end+rem.start-1)/2;
		    		   
		    		   A[pos]=count+1;
		    		   
		    		   if(pos-1>=rem.start)
		    		   PQ.add(new Pair(rem.start,pos-1));
		    		   
		    		   if(pos+1<=rem.end)
		    		   PQ.add(new Pair(pos+1,rem.end));
		    		
		    		count++;
		    	}
			
		    	for(int x:A) {
		    		System.out.print(x+" ");
		    	}
		    	System.out.println();
		    }
	   }
	}