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 q4 {
	public static int count;
	public static class sd implements Comparable<sd>{
		int s;
		int e;
		public sd(int s,int e) {
			this.s=s;
			this.e=e;
		}
		

		@Override
		public int compareTo(sd arg) {
			int l1=this.e-this.s+1;
			int l2=arg.e-arg.s+1;
			if(l1==l2) {
				return this.s-arg.s;
			}
			return l2-l1;
			
		}			
		
	}
	public static void solve(int s,int e,int[] arr) {
		PriorityQueue<sd>order=new PriorityQueue<q4.sd>();
		sd first=new sd(s,e);
		order.add(first);
		while(!order.isEmpty()) {
			sd curr=order.remove();
			if(curr.s>curr.e) {
				continue;
			}
			 s=curr.s;
			 e=curr.e;
			int mid=(s+e)/2;
			arr[mid]=count++;
			if((e-s+1)%2==0) {
				// add right
				sd right=new sd(mid+1,e);
				order.add(right);
				// add left
				sd left=new sd(s,mid-1);
				order.add(left);
			}
			else {
				// add left
				sd left=new sd(s,mid-1);
				order.add(left);
				// add right
				sd right=new sd(mid+1,e);
				order.add(right);
			}
		}
		
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s=new Scanner(System.in);
		int t=s.nextInt();
		for(int i=0;i<t;i++) {
			int n=s.nextInt();
			int[] arr=new int[n];
			count=1;
			solve(0,n-1,arr);
			for(int j=0;j<n;j++) {
				System.out.print(arr[j]+" ");
			}
			System.out.println();
		}

	}

}