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.util.*;

public class TaskA {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        
        int t = Integer.parseInt(br.readLine());
        for (int tt = 0; tt < t; tt++) {
            int n = Integer.parseInt(br.readLine());
            int[] res = solve(n);
            for (int x : res) pw.print(x + " ");
            pw.println();
        }
        
        pw.close();
    }

    private static int[] solve(int n) {
        int[] arr = new int[n];

        PriorityQueue<Pair> q = new PriorityQueue<>(new Comparator<Pair>() {
            @Override
            public int compare(Pair o1, Pair o2) {
                int diff = (o1.b - o1.a) - (o2.b - o2.a);
                if (diff != 0) {
                    return -diff;
                }
                return (o1.a - o2.a);
            }
        });
        q.add(new Pair(0, n - 1));
        int i = 0;
        while (!q.isEmpty()) {
            Pair head = q.poll();
            int l = head.a;
            int r = head.b;

            int m = l + (r - l) / 2;


            if (l < 0 || r >= n || m >= n) continue;
            if (arr[m] != 0) continue;

            i++;

            arr[m] = i;

            q.add(new Pair(m + 1, r));
            q.add(new Pair(l, m - 1));
        }

        return arr;
    }

    static class Pair {
        int a, b;
        public Pair(int x, int y) {
            a = x;
            b = y;
        }

        @Override
        public String toString() {
            return "Pair{" +
                    "a=" + a +
                    ", b=" + b +
                    '}';
        }
    }
}