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.*;
import java.util.Comparator;
import java.util.PriorityQueue;

public class Main{
    static class SubArray  {
        public int L;
        public int R;
        public int size;

        public SubArray(int l,int r){
            L=l;
            R=r;
            size=r-l+1;
        }


    }

    public static void main(String[] args) {
        int test;
        Scanner sc=new Scanner(System.in);
        test=sc.nextInt();
        while(test-->0) {
            int n;
            n=sc.nextInt();
            int arr[] = new int[n+1];

            PriorityQueue<SubArray> pq = new PriorityQueue<SubArray>(5, new Comparator<SubArray>() {
                @Override
                public int compare(SubArray ob1, SubArray ob2) {
                    if (ob1.size == ob2.size)
                        return ob1.L - ob2.L;
                    else
                        return ob2.size - ob1.size;

                }
            });
            pq.add(new SubArray(1, n));
            int k=1;
            for (int i = 1; i <= n; i++) {
                    SubArray ok=pq.poll();
                    int left=ok.L;
                    int right=ok.R;
                    int mid=left+(right-left)/2;
                    arr[mid]=k;
                    k++;
                    pq.add(new SubArray(left,mid-1));
                    pq.add(new SubArray(mid+1,right));
            }
            for(int i = 1; i <= n; i++)
            {
                System.out.print(arr[i] + " ");
            }
            System.out.println();
        }
    }
}