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.awt.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.StringTokenizer;

public class Recursion {
    static long dp[][];
    static int n;
    static int a[];
    static int ans[];
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int pandey=sc.nextInt();
        while (pandey-->0) {
            int n = sc.nextInt();
            ans = new int[n];
            int x = 1;
            PriorityQueue<Point> pinky=new PriorityQueue<>(new Comparator<Point>() {
                @Override
                public int compare(Point o1, Point o2) {
                    if ((o1.y-o1.x+1)-(o2.y-o2.x+1)==0)return o1.x-o2.x;
                    return -((o1.y-o1.x+1)-(o2.y-o2.x+1));
                }
            });
            pinky.add(new Point(0,n-1));
            while (!pinky.isEmpty()){
                Point p=pinky.poll();
                int c=p.x+(p.y-p.x)/2;
                ans[c]=x++;
                if (p.x<=c-1){
                    pinky.add(new Point(p.x,c-1));
                }
                if (c+1<=p.y){
                    pinky.add(new Point(c+1,p.y));
                }
            }
            StringBuilder arun=new StringBuilder();
            for (int tt:ans)arun.append(tt+" ");
            System.out.println(arun);
        }
    }}