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.util.*;
import java.io.*;
import java.text.*;
import java.math.*;
import java.awt.Point;

 public class MEXUM{

	long MOD = 998244353;
	void pre() throws Exception{}
	void solve(int TC) throws Exception{

        int i,j;
        int n = ni();
        for(i=1;i<=n;i++)
        p(i+" ");
        pn("");

    
	}
	
	

	//SOLUTION EtND
	void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
	DecimalFormat df = new DecimalFormat("0.00000000000");
	static boolean multipleTC = true;
	FastReader in;PrintWriter out;
	void run() throws Exception{
	    in = new FastReader();
	    out = new PrintWriter(System.out);
	    //Solution Credits: Taranpreet Singh
	    int T = (multipleTC)?ni():1;
	    pre();for(int t = 1; t<= T; t++)solve(t);
	    out.flush();
	    out.close();
	}
	public static void main(String[] args) throws Exception{
	    new MEXUM().run();
	}
	int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
	void p(Object o){out.print(o);}
	void pn(Object o){out.println(o);}
	void pni(Object o){out.println(o);out.flush();}
	String n()throws Exception{return in.next();}
	String nln()throws Exception{return in.nextLine();}
	int ni()throws Exception{return Integer.parseInt(in.next());}
	long nl()throws Exception{return Long.parseLong(in.next());}
	double nd()throws Exception{return Double.parseDouble(in.next());}

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

	    public FastReader(String s) throws Exception{
	        br = new BufferedReader(new FileReader(s));
	    }

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

	    String nextLine() throws Exception{
	        String str = "";
	        try{   
	            str = br.readLine();
	        }catch (IOException e){
	            throw new Exception(e.toString());
	        }  
	        return str;
	    }
	}
}