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.*;

public class Main {

    public Main() {
    }

    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }

    private void solve() {

        Scanner sc = new Scanner( System.in);
        int tests = sc.nextInt();

        while( tests > 0 ) {
            solution(sc, tests);
            tests--;
        }
        sc.close();
    }

    TreeMap<Integer, ArrayList<Integer>> myMap;
    private void solution( Scanner sc, int testCase) {

        myMap = new TreeMap<Integer, ArrayList<Integer>>( Collections.reverseOrder());
        int n = sc.nextInt();

        rec(1, n);
        int save[] = new int[n];

        //System.out.println( myMap.size());

        int counter = 1;
        for( Map.Entry<Integer, ArrayList<Integer>> entry: myMap.entrySet()) {
            //System.out.print( entry.getKey() +" ");

            Collections.sort( entry.getValue() );

            for( int i: entry.getValue()) save[i-1] = counter++;
        }

        for( int i=0; i<n; i++) System.out.print( save[i] +" ");
        System.out.println();
    }

    private void rec( int i, int j) {

        if( i>j) return;

        //System.out.println("Pos: "+i+" "+j);
        int len =  (j-i+1);
        if( len % 2 == 1 ) {

            if( !myMap.containsKey(len) ) myMap.put(len, new ArrayList<Integer>());
            myMap.get(len).add( (j+i)/2 );
            rec( i, ((j+i)/2)-1 );
            rec( ((j+i)/2)+1 ,j);
        }
        else{

            if( !myMap.containsKey(len) ) myMap.put(len, new ArrayList<Integer>());
            myMap.get( len ).add( (j+i-1)/2 );
            rec( i, ((j+i-1)/2)-1 );
            rec( ((j+i-1)/2)+1 ,j);

        }
    }
}