Problem: You have a sequence a with n elements 1, 2, 3, ..., k - 1, k, k - 1, k - 2, ..., k - (n - k) (k ≤ n < 2k).

Let's call as inversion in a a pair of indices i < j such that a[i] > a[j].

Suppose, you have some permutation p of size k and you build a sequence b of size n in the following manner: b[i] = p[a[i]].

Your goal is to find such permutation p that the total number of inversions in b doesn't exceed the total number of inversions in a, and b is lexicographically maximum.

Small reminder: the sequence of k integers is called a permutation if it contains all integers from 1 to k exactly once.

Another small reminder: a sequence s is lexicographically smaller than another sequence t, if either s is a prefix of t, or for the first i such that s_i ≠ t_i, s_i < t_i holds (in the first position that these sequences are different, s has smaller number than t).

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first and only line of each test case contains two integers n and k (k ≤ n < 2k; 1 ≤ k ≤ 10^5) — the length of the sequence a and its maximum.

It's guaranteed that the total sum of k over test cases doesn't exceed 10^5.

Output

For each test case, print k integers — the permutation p which maximizes b lexicographically without increasing the total number of inversions.

It can be proven that p exists and is unique.

Example

Input


4
1 1
2 2
3 2
4 3


Output


1 
1 2 
2 1 
1 3 2 

Note

In the first test case, the sequence a = [1], there is only one permutation p = [1].

In the second test case, the sequence a = [1, 2]. There is no inversion in a, so there is only one permutation p = [1, 2] which doesn't increase the number of inversions.

In the third test case, a = [1, 2, 1] and has 1 inversion. If we use p = [2, 1], then b = [p[a[1]], p[a[2]], p[a[3]]] = [2, 1, 2] and also has 1 inversion.

In the fourth test case, a = [1, 2, 3, 2], and since p = [1, 3, 2] then b = [1, 3, 2, 3]. Both a and b have 1 inversion and b is the lexicographically maximum.

Solution (in JAVA):

Solution:
/*input
4
1 1
2 2
3 2
4 3

*/
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
	static PrintWriter out;
	static int MOD = 1000000007;
	static FastReader scan;

    /*-------- I/O using short named function ---------*/
	public static String ns(){return scan.next();}
	public static int ni(){return scan.nextInt();}
	public static long nl(){return scan.nextLong();}
	public static double nd(){return scan.nextDouble();}
	public static String nln(){return scan.nextLine();}
	public static void p(Object o){out.print(o);}
	public static void ps(Object o){out.print(o + " ");}
	public static void pn(Object o){out.println(o);}
    /*-------- for output of an array ---------------------*/
     static void iPA(int arr []){
     	 StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void lPA(long arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void sPA(String arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void dPA(double arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
    /*-------------- for input in an array ---------------------*/
	static void iIA(int arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = ni();
	}
	static void lIA(long arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = nl(); 
	}
	static void sIA(String arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = ns(); 
	}
	static void dIA(double arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = nd();
	}
	/*------------ for taking input faster ----------------*/
     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;
		}
	}

	// Method to check if x is power of 2
    static boolean isPowerOfTwo (int x) { return x!=0 && ((x&(x-1)) == 0);}
    //Method to return lcm of two numbers
    static int gcd(int a, int b){return a==0?b:gcd(b % a, a); }
    // method to return LCM of two numbers
    static int lcm(int a, int b){return (a / gcd(a, b)) * b;}
    //Method to count digit of a number
    static int countDigit(long n){return (int)Math.floor(Math.log10(n) + 1);}

    //Method to find the max in an array
    static int getMax(int arr[]){
    	int max = arr[0];
    	for(int i=0; i<arr.length; i++){
    		max = arr[i]>max?arr[i]:max;
    	}
    	return max;
    }
    //Method to find the min in an array
    static int getMin(int arr[]){
    	int min = arr[0];
    	for(int i=0; i<arr.length; i++){
    		min = min>arr[i]?arr[i]:min;
    	}
    	return min;
    }
    
	

	public static void main (String[] args) throws java.lang.Exception
	{
		OutputStream outputStream =System.out;
        out =new PrintWriter(outputStream);
        scan =new FastReader();
		//for fast output sometimes
		StringBuilder sb = new StringBuilder();
		int t = ni();
		while(t-->0){
			int n = ni(), k = ni();
			int arr[] = new int[n];
			int j=0;
			HashSet<Integer> hash = new HashSet<Integer>();
			for(int i=1; i<=k; i++){
				arr[j++] = i;
			}
			int m = k;
			for(int i=k+1; i<=n; i++){
				arr[j++] = --m;
			}
			//iPA(arr);
			//pn("");
			ArrayList<Integer> list = new ArrayList<Integer>();
			for(int i=n-1; i>=0; i--){
				if(!hash.contains(arr[i])){
					hash.add(arr[i] );
					list.add(arr[i]);
				}
			}
			for(int i=list.size()-1; i>=0; i--){
				sb.append(list.get(i) + " ");
			}
			sb.append("\n");
        }
        pn(sb);
		out.close();
	}
	
}