A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) — 

  * (p_i  OR  p_{i+1}  OR  …  OR  p_{j-1}  OR  p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)



In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray. 

Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).

Output

For every test, output any good permutation of length n on a separate line. 

Example

Input


3
1
3
7


Output


1
3 1 2
4 3 5 2 7 1 6

Note

For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below. 

  * 3 OR 1 = 3 ≥ 2 (i = 1,j = 2) 
  * 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3) 
  * 1 OR 2 = 3 ≥ 2 (i = 2,j = 3) 
  * 1 ≥ 1 (i = 2,j = 2) 



Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
import java.io.*;
import java.util.*;

public class A {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastScanner sc = new FastScanner();
	    out = new PrintWriter(new BufferedOutputStream(System.out));
	     
	    int t=sc.nextInt();
	    while(t-->0) {
	    	int n=sc.nextInt();
	    	if(n==1) {
	    		out.println(1);
	    	}
	    	else if(n==2) {
	    		out.println("1 2");
	    	}
	    	else {
		    	int[] array=new int[n];
		    	int count=1;
		    	for(int i=n-2;i>=0;i=i-2) {
		    		array[i]=count;
		    		count++;
		    	}
		    	if(n%2==0) {
		    		for(int i=1;i<n-3;i=i+2) {
		    			array[i]=count;
		    			count++;
		    		}
		    	}
		    	else {
		    		for(int i=0;i<n-3;i=i+2) {
			    		array[i]=count;
			    		count++;
			    	}
		    	}
		    	array[n-1]=count;
		    	array[n-3]=count+1;
		    	for(int i=0;i<n;i++) {
		    		out.print(array[i]+ " ");
		    	}
		    	out.println();
	    	}
	    }
	    out.close();
	}

	   public static PrintWriter out;
	      
	   public static class FastScanner {
	      BufferedReader br;
	      StringTokenizer st;
	 
	      public FastScanner() {
	         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;
	      }

	   }

}
