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 com.sun.org.apache.bcel.internal.generic.ALOAD;
import java.util.*;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import javafx.util.Pair;
import jdk.nashorn.internal.ir.BreakNode;

public class d {

    static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
    static StringBuilder out = new StringBuilder();
    static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
    static PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));

    static int I(String s) {
        return Integer.parseInt(s);
    }

    static long L(String s) {
        return Long.parseLong(s);
    }

    static double D(String s) {
        return Double.parseDouble(s);
    }

    static String[] SA() throws IOException {
        return bf.readLine().split(" ");
    }

    public static void main(String[] args) throws IOException {
        int t = I(bf.readLine());

        while (t-- > 0) {
            int n = I(bf.readLine());

            al = new ArrayList<>();
            for (int i = 1; i <= n + 1; i++) {
                al.add(new ArrayList<>());
            }
            rec(0, n-1, n);
            int a[]=new int[n];
            int inc=1;
            for (int i = n; i >=1; i--) {
                Collections.sort(al.get(i));
                for (Integer x : al.get(i)) {
                   if(a[x]==0){// System.out.println(x+"  "+inc+"  "+i);
                   a[x]=inc;
                    inc++;
                }}
               
            }
           // System.out.println(inc);
for(int i=0;i<n;i++){
if(a[i]>0){out.append(a[i]+" ");}
else{out.append(inc+" ");inc++;}

}out.append("\n");
        }

       System.out.print(out);

    }

    static ArrayList<ArrayList<Integer>> al;

    static void rec(int l, int r, int n) {
    //    System.out.println(l + " " + r);
        
        if ((l + r) % 2 == 1) {
            int mid = ((l + r-1) / 2) ;

            al.get(r-l).add(mid);
            al.get(r-l).add(mid);
        if (mid<r){  rec(mid+1, r, n);}
         if(l<mid) { rec(l, mid-1, n);
          }
        } else {
            int mid = ((l + r) / 2);
             al.get(r-l).add(mid);
            al.get(r-l).add(mid);
          if(mid<r){  rec(mid+1, r, n);}
         if(l<mid) { rec(l, mid-1, n);
          }
        }
       if(l==r||r==0||l==n-2||l+1==r){return;}
    }
}
/*
in:
6
1
2
3
4
5
6
otu:
1 
1 2 
2 1 3 
3 1 2 4 
2 4 1 3 5 
3 4 1 5 2 6 
*/