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.*;

public class Main {
  public static void main(String[] args) throws IOException {
    FastReader in = new FastReader();
    PrintWriter out = new PrintWriter(System.out);

    Task solver = new Task();
    solver.solve(in, out);

    out.flush();
    out.close();
  }

  static class Task {

    // Consts
    final int MOD = (int)1e9+7;
    final int MAXN = (int)2e5+2;
    final int INF = 0x3f3f3f3f;

    // G & Funcs
    int a[] = new int[MAXN];

    class Pair implements Comparable<Pair>{
      private int l;
      private int r;

      Pair(int l, int r) {
        this.l = l;
        this.r = r;
      }

      public int getL() {
        return l;
      }

      public int getR() {
        return r;
      }

      int getRangeSize() {
        return r-l+1;
      }

      int getMid() {
        if((getRangeSize()&1) == 0)
          return (l+r-1)/2;
        return (l+r)/2;
      }


      @Override
      public int compareTo(Pair o) {
        int cmp = o.getRangeSize() - getRangeSize();
        if(cmp == 0)
          cmp = l - o.l;

        return cmp;
      }
    }


    void solve(FastReader in, PrintWriter out) throws IOException {

      int T;
      T = in.nextInt();

      Queue<Pair> q = new PriorityQueue<>();
      Pair curr, l, r;

      int n;
      while(T-- != 0) {
        n = in.nextInt();

        q.add(new Pair(1, n));
        for(int i=1;i<=n;++i) {
          curr = q.poll();
          a[curr.getMid()] = i;

          l = new Pair(curr.getL(), curr.getMid()-1);
          r = new Pair(curr.getMid()+1, curr.getR());

          if(l.getRangeSize() > 0)
            q.add(l);
          if(r.getRangeSize() > 0)
            q.add(r);

        }

        for(int i=1;i<=n;++i)
          out.print(a[i]+" ");
        out.println();

      }
    }
  }

  static class FastReader {

    BufferedReader br;
    StringTokenizer st;

    public FastReader(InputStream is) {
      br = new BufferedReader(new InputStreamReader(is), 32768);
    }

    public FastReader() {
      this(System.in);
    }

    public String next() {
      if (st == null || !st.hasMoreTokens()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (Exception e) {
        }
      }
      return st.nextToken();
    }

    public int nextInt() {
      return Integer.parseInt(next());
    }

    public long nextLong() {
      return Long.parseLong(next());
    }

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

    public String nextLine() {
      try {
        return br.readLine();
      } catch (Exception e) {
      }
      return null;
    }

    public boolean hasNext() throws IOException {
      if (st != null && st.hasMoreTokens()) {
        return true;
      }
      String s = br.readLine();
      if (s == null || s.isEmpty()) {
        return false;
      }
      st = new StringTokenizer(s);
      return true;
    }
  }
}