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.lang.*;
import java.io.*;
 
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{   	public static void main (String[] args)throws IOException {
    try{
	   //BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
	   Scanner sc = new Scanner(System.in);
	  // int t=Integer.parseInt(br.readLine());
	   int t=sc.nextInt();
	// int t=1;
	   while(t-->0){
	 //int n=Integer.parseInt(br.readLine())
	int n = sc.nextInt();
	//int k = sc.nextInt();
	//long d=Long.parseLong(br.readLine());
	//String[] s= br.readLine().trim().split(" ");
	// Map<Integer,Integer> m = new TreeMap<Integer,Integer>();
 List<Integer> f = new ArrayList<Integer>();
 //List<Integer> s = new ArrayList<Integer>();
	 //<Integer> v= new Vector<Integer>();
	 //Set<Integer> s= new HashSet<Integer>();
	 int a[] = new int[n+1];
	// int b[] = new int[n];
  PriorityQueue<pair> pq = new	PriorityQueue<pair>(new segmentComparator()); 
			pq.add(new pair(1,n));
			int l=0,r=0;
			for(int i=1;i<=n;i++){
			   // System.out.println(pq);
			   pair p = pq.poll(); 
			   l=p.x;
			   r=p.y;
			    int mid = (l+r)/2;
			    a[mid]=i;
			    pq.add(new pair(l,mid-1));
			    pq.add(new pair(mid+1,r));
			}
			for(int i=1;i<=n;i++) System.out.print(a[i]+" ");
			System.out.println();
}	
}
catch(Exception e){
    
}
	}}
	
	class pair{
	    int x;
	    int y;
	    pair(int a, int b){
	        x=a;
	        y=b;
	    }
	}
	
	class segmentComparator implements Comparator<pair>{
	    public int compare(pair p1, pair p2){
	         int length1 =p1.y-p1.x;
	        int length2 = p2.y-p2.x;
	        if(length1>length2) return -1;
	       else if(length1<length2) return 1;
	       else{
	           return (p1.x<p2.x) ? -1 : 1; 
	       }
	    }
	}