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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.sql.Array;
import java.util.*;

public class D {
    static StringTokenizer st;
    static BufferedReader br;
    static PrintWriter pw;

    static String nextToken() {
        try {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine());
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
        return st.nextToken();
    }

    static int nextInt() {
        return Integer.parseInt(nextToken());
    }

    static long nextLong() {
        return Long.parseLong(nextToken());
    }

    static String nextLine() {
        try {
            return br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }

    public static void main(String[] args) {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(System.out);
        run();
        pw.close();
    }

    private static void run() {
        int t = nextInt();
        for (int i = 0; i < t; i++) {
            int n = nextInt();
            TreeSet<Pair> pairs = new TreeSet<>();
            pairs.add(new Pair(1, n));
            int[] ans = new int[n];
            int k = 1;
            while (pairs.size() > 0) {
                Pair p = pairs.first();
                pairs.remove(p);
                int mid = (p.r + p.l) / 2;
                Pair[] ps = new Pair[2];
                ps[0] = new Pair(p.l, mid - 1);
                ps[1] = new Pair(mid + 1, p.r);
                for (int j = 0; j < 2; j++) {
                    if (ps[j].size > 0) {
                        pairs.add(ps[j]);
                    }
                }
                ans[mid - 1] = k++;
            }
            for (int j = 0; j < n; j++) {
                pw.print(ans[j] + " ");
            }
            pw.println();
        }
    }

    static class Pair implements Comparable<Pair> {
        int l;
        int r;
        int size;

        public Pair(int l, int r) {
            this.l = l;
            this.r = r;
            this.size = r - l + 1;
        }

        @Override
        public int compareTo(Pair o) {
            if (o.size != size) return Integer.compare(o.size, size);
            return Integer.compare(l, o.l);
        }
    }
}