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.PriorityQueue;
import java.util.Scanner;

public class r642a {
	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		int t=scan.nextInt();
		for(int tt=0;tt<t;tt++) {
			int n=scan.nextInt();
			PriorityQueue<interval> q=new PriorityQueue<>();
			q.offer(new interval(0,n-1));
			int[] res=new int[n];
			int ct=1;
			
			while(!q.isEmpty()) {
//				System.out.println(q);
				interval cur=q.poll();
				int mid=cur.middle();
				
				res[mid]=ct++;
				
				if(cur.l<=mid-1) {
					q.offer(new interval(cur.l,mid-1));
				}
				if(mid+1<=cur.r) {
					q.offer(new interval(mid+1,cur.r));
				}
			}
			for(int i:res) System.out.print(i+" ");
			System.out.println();
		}
	}
	static class interval implements Comparable<interval> {
		int l,r;
		interval(int l, int r) {
			this.l=l;
			this.r=r;
		}
		int middle() {
			return (l+r)/2;
		}
		@Override
		public int compareTo(interval o) {
			if(r-l==o.r-o.l) { 
				return l-o.l;
			}
			return (o.r-o.l)-(r-l);
		}
		public String toString() {
			return l+" "+r;
		}
	}
}