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 javafx.concurrent.Task;
import sun.awt.image.ImageWatched;
import sun.reflect.generics.tree.Tree;

import javax.jws.soap.SOAPBinding;
import java.util.*;

public final class Main {


    static class Length implements Comparable<Length>{
        @Override
        public int compareTo(Length o) {

            int len1 = r - l + 1;

            int len2 = o.r - o.l + 1;

            if (len1 == len2) return ((Integer) l).compareTo(o.l);

            else return ((Integer) len2).compareTo(len1);
        }

        public int mid() {

            return (l+r)/2;
        }


        int l, r;

        public Length(int l, int r) {

            this.l = l;

            this.r =r;

        }


    }
    public static void main(String[] args) {

        Scanner s  = new Scanner(System.in);

        int TC =s.nextInt();

        out: for (int tc = 0; tc < TC; tc++) {

            int n = s.nextInt();

            int[] arr = new int[n];

            int i = 1;

            PriorityQueue<Length> q = new PriorityQueue<>();

            q.add(new Length(0, n-1));

            while (q.isEmpty() == false) {

                Length l = q.poll();

                int left = l.l;

                int right = l.r;

                int mid = l.mid();


                arr[mid] = i;


                   if (mid-1 >= left) q.add(new Length(left, mid - 1));

                   if (mid+1 <= right) q.add(new Length(mid + 1, right));


                i++;
            }

            for (int k : arr) {
                System.out.print(k + " ");
            }

            System.out.print("\n");
        }

    }
}