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 D1353 {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	static StringTokenizer st;
	public static void main(String[]args)throws IOException {
		int t = readInt();
		for(int cases = 0; cases<t; cases++) {
			int n = readInt();//base 10^6 first is sublength second is index
			PriorityQueue<Long>indices = new PriorityQueue<Long>(Collections.reverseOrder());
			indices.add((long)n*1000000+(-1)*1);
			int[]array = new int[n+1];
			for(int i = 1; i<=n; i++) {
				long num = indices.poll();
				int len = (int)(num/1000000+1);
				int l = (int)(1000000-num%1000000);
				long r = l+(len-1);
				if((r-l+1)%2==1)array[(int)(l+r)/2] = i;
				else array[(int)(l+r-1)/2] = i;
				if(len==1) {
					
				}
				else if(len%2==1) {
					int l1 = l;
					int l2 = l+((len+1)/2);
					indices.add((long)(len/2)*1000000 + (-1)*l1);
					indices.add((long)(len/2)*1000000 + (-1)*l2);
				}
				else if(len==2){
					indices.add((long)(len/2)*1000000 +(-1)*(l+1));
				}
				else {
					int l1 = l;
					int l2 = l+(len/2);
					indices.add((long)(len/2-1)*1000000 +(-1)*l1);
					indices.add((long)(len/2)*1000000 +(-1)*l2);
				}
			}
			for(int i = 1; i<=n; i++) {
				pr.print(array[i]);
				if(i!=n)pr.print(" ");
			}
			pr.println();
		}
		pr.close();
	}
	static String next () throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine().trim());
		return st.nextToken();
	}
	static long readLong () throws IOException {
		return Long.parseLong(next());
	}
	static int readInt () throws IOException {
		return Integer.parseInt(next());
	}
	static double readDouble () throws IOException {
		return Double.parseDouble(next());
	}
	static char readCharacter () throws IOException {
		return next().charAt(0);
	}
	static String readLine () throws IOException {
		return br.readLine().trim();
	}
}