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.math.BigInteger;
import java.util.*;

import java.util.ArrayList;

import java.util.Scanner;


public class adaking {

    static class pair {
        int first;
        int second;

        pair(int a, int b) {
            first = a;
            second = b;
        }
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                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());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static final int mod = 1000000007;

    public static int gcd(int a, int b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        sc.nextLine();

        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int a[]=new int[n];

            int[] res = new int[n + 1];
            PriorityQueue<pair> q = new PriorityQueue<>(new Comparator<pair>() {
                @Override
                public int compare(pair o1, pair o2) {
                    if(o1.second-o1.first>o2.second-o2.first)
                        return -1;
                    else if(o1.second-o1.first<o2.second-o2.first)
                        return 1;
                    else
                    return o1.first-o2.first;
                }
            });
            q.add(new pair(1,n));int count=1;
            while(!q.isEmpty())
            {
                pair g=q.poll();
                int e=g.first;
                int f=g.second;int mi=0;
                if(e-f+1%2!=0)
                    mi=(e+f)/2;
                else
                    mi=(e+f-1)/2;
                a[mi-1]=count++;
                if(count>n)
                    break;

                if(mi+1>=1&&mi-1<=n&&f<=n&&f>=1)
                    q.add(new pair(mi+1,f));
                if(mi-1>=1&&mi-1<=n&&e>=1&&e<=n)
                    q.add(new pair(e,mi-1));

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