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:
/*
 *created by Kraken on 14-05-2020 at 19:50
 */
//package com.kraken.cf.cf642;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Comparator;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.StringTokenizer;

public class D {

  private static class Range {
    int l, r;

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

    @Override
    public String toString() {
      return "Range{" +
        "l=" + l +
        ", r=" + r +
        '}';
    }
  }

  public static void main(String[] args) {
    FastReader sc = new FastReader();
    int t = sc.nextInt();
    while (t-- > 0) {
      int n = sc.nextInt();
      PriorityQueue<Range> q = new PriorityQueue<>(new Comparator<Range>() {

        @Override
        public int compare(Range o1, Range o2) {
          int d1 = o1.r - o1.l + 1;
          int d2 = o2.r - o2.l + 1;
          if (d1 != d2) {
            return d2 - d1;
          } else {
            return o1.l - o2.l;
          }
        }
      });
      int[] res = new int[n + 1];
      int cnt = 1;
      q.add(new Range(1, n));
      while (!q.isEmpty()) {
        Range r = q.poll();
        int v = r.r - r.l + 1;
        int mid;
        if (v % 2 != 0) mid = (r.l + r.r) / 2;
        else mid = (r.l + r.r - 1) / 2;
        res[mid] = cnt++;
        if (mid > r.l) q.add(new Range(r.l, mid - 1));
        if (r.r > mid) q.add(new Range(mid + 1, r.r));
      }
      StringBuilder sb = new StringBuilder();
      for (int i = 1; i <= n; i++) sb.append(res[i]).append(" ");
      System.out.println(sb);
    }
  }

  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;
    }
  }
}