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:
//package div3._1353;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.util.stream.Collectors;

public class ConstructingTheArray {

    private final FastReader fr = new FastReader();

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

    private void solve() {
        int t = fr.nextInt();
        while (t-- > 0) {
            int n = fr.nextInt();
            int[] arr = new int[n];

            PriorityQueue<Segment> segments = new PriorityQueue<>();

            segments.add(new Segment(1, n));

            int index;
            for (int i = 1; !segments.isEmpty(); i++) {
                Segment p = segments.poll();
                if ((p.right - p.left + 1) % 2 == 0)
                    index = (p.left + p.right - 1) / 2;
                else
                    index = (p.left + p.right) / 2;

                arr[index - 1] = i;
                if (p.left < index)
                    segments.add(new Segment(p.left, index - 1));
                if (p.right > index)
                    segments.add(new Segment(index + 1, p.right));
            }

            System.out.println(
                    Arrays.stream(arr)
                            .mapToObj(String::valueOf)
                            .collect(Collectors.joining(" "))
            );
        }
    }

    static class Segment implements Comparable<Segment> {
        private final int size;
        private final int left;
        private final int right;

        Segment(int left, int right) {
            this.left = left;
            this.right = right;
            this.size = right - left;
        }

        public int compareTo(Segment other) {
            if (this.size == other.size)
                return this.left - other.left;
            return other.size - this.size;
        }
    }

    static class FastReader {
        private final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        private StringTokenizer st;

        public String nextLine() {
            try {
                return br.readLine();
            } catch (IOException ex) {
                throw new RuntimeException(ex);
            }
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(nextLine());
            }
            return st.nextToken();
        }

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

}