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.io.*;
import java.util.PriorityQueue;
 
public class D
{
    static class Pair implements Comparable<Pair>
    {
        int l, r;
        Pair(int l, int r){this.l = l;this.r = r;}
        public int compareTo(Pair b)
        {
            if(Integer.compare(b.r-b.l+1,this.r-this.l+1)==0)
                return Integer.compare(this.l,b.l);
            return (Integer.compare(b.r-b.l+1,this.r-this.l+1));
        }
    }
    public static void main(String[] args) throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
 
        int i,N;
 
        int T=Integer.parseInt(br.readLine().trim());
        StringBuilder sb=new StringBuilder();
 
        while(T-->0)
        {
            N=Integer.parseInt(br.readLine().trim());
            int a[]=new int[N];
 
            PriorityQueue<Pair> PQ=new PriorityQueue<>();
            PQ.add(new Pair(0,N-1));
 
            for(i=1;i<=N;i++)
            {
                Pair tmp=PQ.poll();
                int pos=(tmp.l+tmp.r)/2;
                a[pos]=i;
 
                int l1=tmp.l,r1=pos-1,l2=pos+1,r2=tmp.r;
                if(l1<=r1) PQ.add(new Pair(l1,r1));
                if(l2<=r2) PQ.add(new Pair(l2,r2));
            }
            for(i=0;i<N;i++) sb.append(a[i]).append(" ");
            sb.append("\n");
        }
        System.out.println(sb);
    }
}