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.*;
public class p1 {
	static ArrayList<Node> list  = new ArrayList<>();
	static void solve(int l,int r,int n){
		if(l>r)return;
		if(list.size()==n)return;
		int ind = 0;
		if((r-l+1)%2==1)ind = l+(r-l)/2;
        else ind = l+ (r-l-1)/2;
		list.add(new Node(r-l+1,l,r,ind));

		solve(l,ind-1,n);
		if(list.size()==n)return;
		solve(ind+1,r,n);
		if(list.size()==n)return;

	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t--!=0){
			list = new ArrayList<Node>();
			int n = sc.nextInt();
			int a[]  = new int[n];
			solve(0,n-1,n);
			Collections.sort(list,new sort());
			for(int i=0;i<n;i++){
				a[list.get(i).i] = i+1;
			}
			for(int i=0;i<n;i++)System.out.print(a[i]+" ");
			System.out.println();
		}
	}
}
class Node{
	int d,l,r,i;
	public Node(int d,int l,int r,int i){
		this.d = d;
		this.l = l;
		this.r = r;
		this.i = i;
	}
}
class sort implements Comparator<Node>{
	public int compare(Node a,Node b){
		if(a.d!=b.d){
			return b.d - a.d;
		}
		return a.l - b.l;
	}
}