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.*;

public class D{
    public static void main(String[] args) {
        FastReader fs=new FastReader();
        PrintWriter out=new PrintWriter(System.out);
        int t=1;
        t=fs.nextInt();
        for(int i=1;i<=t;i++) {
        	D m=new D();
            m.solve(i,out,fs);
        }
        out.close();
    } 
   
    public void solve(int test,PrintWriter out,FastReader fs){
        int n=fs.nextInt();
        PriorityQueue<Seg> pq=new PriorityQueue<Seg>((u,v)->{
            if( (u.r-u.l)==(v.r-v.l))return u.r-v.r;
            return (v.r-v.l)-(u.r-u.l);
        });
        int ans[]=new int[n+1];
        pq.add(new Seg(1,n));
        int pos=0;
        while(!pq.isEmpty()){
            Seg top=pq.poll();
            //out.println(top.l+" "+top.r);
            if(top.l>top.r)continue;
            if(top.l<top.r){
                int mid=(top.l+top.r)/2;
                pq.add(new Seg(top.l,mid-1));
                pq.add(new Seg(mid+1,top.r));
            }
            ans[ (top.l+top.r)/2 ] =++pos;
        }
        for(int i=0;i<n;i++){
            out.print(ans[i+1]+" ");
        }
        out.println();
    }
}
class Seg{
    int l,r;
    Seg(int l,int r){
        this.l=l;
        this.r=r;
    }
}
class FastReader{
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st=new StringTokenizer("");  
    String next(){
        while(!st.hasMoreTokens()){
            try {
                st=new StringTokenizer(br.readLine()); 
            } catch (IOException e) {
				e.printStackTrace();
			}
        }
        return st.nextToken();
    }
    int nextInt(){
        return Integer.parseInt(next());
    }
    long nextLong(){
        return Long.parseLong(next());
    }
    int[] nextIntArray(int n){
        int a[]=new int[n];
        for(int i=0;i<n;i++)a[i]=nextInt();
        return a;
    } 
    long[] nextLongArray(int n){
        long a[]=new long[n];
        for(int i=0;i<n;i++)a[i]=nextInt();
        return a;
    }
    char[] nextCharArray(){
        char a[]=next().toCharArray();
        return a;
    }
}