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:
import java.util.*;

import java.io.*;


 public class Codechef {
	 static class FR{
			BufferedReader br;
			StringTokenizer st;
			public FR() {
				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; 
		    } 
		}
	 static int mod  = 1000000007;
	 static int gcd(int a, int b)
	  {      
	     if (b == 0)
	        return a;
	     return gcd(b, a % b); 
	  }
		
	 public static void main(String args[]) {
		 FR sc = new FR();
			int t = sc.nextInt();
			while(t-- > 0) {
				int n = sc.nextInt() , k = sc.nextInt();
				
				int[] arr = new int[n];
				int inv = 1;
				for(int i = 0 ; i<n ;i++) {
					if(i+1<=k) arr[i] = i+1;
					else {
						arr[i]  = i - k + 1;
						inv ++;
					}
				}
				
//				for(int e:arr) {
//					System.out.print(e+" ");
//				}System.out.println();
//				
				int[] p = new int[k];
				for(int i = 0 ;i<k-inv ; i++) {
					p[i] = arr[i];
				}
				int num = k;
				for(int i = k-inv ; i<k;i++) {
					p[i] = num--;
				}
				for(int e:p) {
					System.out.print(e+ " ");
				}System.out.println();
			}
		 
	 }
	
		
  }