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 
import java.io.ObjectInputStream;
import java.lang.reflect.Array;
import java.util.*;

public class Main {


    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int t;
        t = input.nextInt();
        while (t > 0) {
            int n = input.nextInt();
            if (n == 1) {
                System.out.println(1);
                t--;
                continue;
            }
            int[] a = new int[n];
            PriorityQueue<Pair> segments = new PriorityQueue<>();
            segments.add(new Pair(0,n-1));
            int i = 1;
            while (i <= n) {
                Pair current  = segments.peek();
                segments.remove(current);
                int index = (current.right + current.left)/2;
                a[index]=i;
                i++;
                if (index-1>=current.left) segments.add(new Pair(current.left,index-1));
                if (current.right >= index+1) segments.add(new Pair(index+1, current.right));
            }

            for (int k = 0; k < n; k++) {
                System.out.print(a[k] + " ");
            }
            System.out.println();
            //System.out.println(segments.size());
            t--;
        }
    }

}

class Pair implements Comparable<Pair>{
    public int left;
    public int right;

    public Pair(int left, int right){
        this.left = left;
        this.right = right;
    }

    @Override
    public int compareTo(Pair p) {
        if (this.right - this.left > p.right - p.left){
            return -1;
        }
        else if (this.right - this.left < p.right - p.left){
            return 1;
        }
        else{
            if (this.left<p.left)
                return -1;
            else
                return 1;
        }
    }
}




