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.util.StringTokenizer; 
import java.util.Collections;
import java.util.PriorityQueue;
import java.util.Comparator;
public class D {
	public static void main(String args[]) {
		PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>() {
			 
            public int compare(int[] e1, int[] e2) {
                if (e1[1] == e2[1]) {
                    return e1[0] - e2[0];
                } else return e2[1] - e1[1];
            }
        });
		
		FastReader in = new FastReader();
		int t = in.nextInt();
		while(t-->0) {
			int n = in.nextInt();
			int[] arr = new int[n];
			int l = 0, r=n-1;
			int d =1; 
			int k=0;
			while(l<=r) {
				if((r-l+1)%2!=0) {
					k =(l+r)/2;
				}else {
					k = (l+r-1)/2;
				}
				arr[k] = d;
				d++;
				if(r>l){
					if(k-l>0) pq.add(new int[] {l, k-l});
					if(r-k>0) pq.add(new int[] {k+1, r-k});	
				}
				if(pq.size()>0) {
					int[] s = pq.remove();
					l = s[0];
					r= l+s[1]-1;
				}else break;
			}
			for(int num : arr) System.out.print(num+" ");
			System.out.println("");
			pq.clear();
			d=0;
		}
	}
	static class FastReader 
			{ 
				BufferedReader br; 
				StringTokenizer st; 

				public FastReader() 
				{ 
					br = new BufferedReader(new
							InputStreamReader(System.in)); 
				} 

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

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

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

				double nextDouble() 
				{ 
					return Double.parseDouble(next()); 
				} 

				String nextLine() 
				{ 
					String str = ""; 
				try
				{ 
					str = br.readLine(); 
				} 
				catch (IOException e) 
				{ 
					e.printStackTrace(); 
				} 
				return str; 
			} 
		}
	
}