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 d {
	static class Pair
	{
		public int a,b;
		public Pair(int a, int b)
		{
			this.a=a;
			this.b=b;
		}
	}
	public static void main(String[] args)throws IOException {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
		int cases = in.nextInt();
		
		for(int t = 0; t < cases; t++)
		{
			int n=in.nextInt();
			int[] ans = new int[n+1];

			PriorityQueue<Pair> pq = new PriorityQueue<Pair>(new Comparator<Pair>()
				{
					@Override
					public int compare(Pair p1, Pair p2)
					{
						int v1 = p1.b-p1.a+1;
						int v2 = p2.b-p2.a+1;
						if(v1==v2)
							return p1.a-p2.a;
						return (p2.b-p2.a+1)-(p1.b-p1.a+1);
					}
				});
			int it = 1; pq.add(new Pair(1,n));
			while(pq.size()!=0)
			{
				Pair p = pq.poll();
				ans[(p.a+p.b)/2]=it;
				int mid = (p.a+p.b)/2;
				// out.println(mid);
				it++;
				if(mid-1>=1 && mid-1>=p.a)
					pq.add(new Pair(p.a,mid-1)); 
				if(mid+1<=n && mid+1<=p.b)
					pq.add(new Pair(mid+1,p.b));
			}
			for(int i=1;i<=n;i++)
				out.print(ans[i]+" ");
			out.println();
		}

        out.close();
	}
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

        public long nextLong() {
            return Long.parseLong(next());
        }

    }

}