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.Comparator;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class EasyCodeforces {

    static class Pair {
        int a;
        int b;

        Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        StringTokenizer stk;

        Comparator<Pair> comp = new Comparator<Pair>() {
            @Override
            public int compare(Pair o1, Pair o2) {
                if ((o1.b - o1.a) < (o2.b - o2.a)) return -1;
                else if ((o1.b - o1.a) > (o2.b - o2.a)) return 1;
                else {
                    if(o1.a < o2.a) return 1;
                    else if(o2.a < o1.a)return -1;
                    return 0;
                }
            }
        };

        int t = Integer.parseInt(in.readLine());
        //System.out.println("testcase : " + t);
        while (t-- > 0) {
            int n = Integer.parseInt(in.readLine());
            PriorityQueue<Pair> pq = new PriorityQueue<>(comp.reversed());
            int[] ans = new int[n];
            pq.add(new Pair(1, n));
            for (int i = 0; i < n; i++) {
                Pair top = pq.poll();
                int l = top.a;
                int r = top.b;
                int mid = (l + r) / 2;
                ans[mid - 1] = i + 1;
                if (mid > l) {
                    pq.add(new Pair(l, mid - 1));
                }
                if (mid < r) {
                    pq.add(new Pair(mid + 1, r));
                }
            }
            for (int i = 0; i < n; i++) {
                out.print(ans[i] + " ");
            }
            out.println();
        }

        out.close();
        in.close();
    }
}