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.HashMap;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Task2 {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		for(int tt=0;tt<t;tt++) {
			int n=sc.nextInt();
			PriorityQueue<Pair> pq=new PriorityQueue<Pair>((p1,p2)->(p2.length-p1.length==0)?p1.l-p2.l:p2.length-p1.length);
			pq.add(new Pair(0,n-1));
			int output[]=new int[n];
			int i=1;
			while(!pq.isEmpty()) {
				Pair p=pq.remove();
				///System.out.println(p.l+" "+p.r);
				int mid=(((p.r-p.l)+1)%2!=0)?p.l+(p.r-p.l)/2:p.l+((p.r-p.l)-1)/2;
				//System.out.println("trenutni interval "+p.l+" "+p.r+" "+mid);
				output[mid]=i;
				if(mid-1>=p.l)
					pq.add(new Pair(p.l,mid-1));
				if(mid+1<=p.r)
					pq.add(new Pair(mid+1,p.r));
				i++;
			}
			for(int k=0;k<n;k++) {
				System.out.print(output[k]+" ");
			}
			System.out.println();
		}
	}
	public static class Pair{
		int l;
		int r;
		int length;
		public Pair(int l,int r) {
			this.l=l;
			this.r=r;
			this.length=(r-l)+1;
		}
	}
}