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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

import java.util.StringTokenizer;
 
public class C_No_More_Inversions {
    static FastReader scanner = new FastReader();

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(
                new InputStreamReader(System.in));
        }
         
        int[] arrayInput(int size)
        {
            int a[]=new int[size];
            for(int i=0;i<size;i++)
            {
                a[i]=nextInt();
            }
            return a;
        }
 
        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 void forLoop(int arr[],int start,int last,int value)
    {
        for(int i=start;i<last;i++)
         arr[i]=value;
    }
     static void forLoopEndEqual(int arr[],int start,int last,int value)
    {
        for(int i=start;i<=last;i++)
         arr[i]=value;
    }
    static void debug(int a){
        System.out.print(a+" ");
    }
     static void debug(int a,int b){
        System.out.print(a+" "+b);
    }
     static void lineBreak(){
        System.out.println("");
    }

    
  /////////////////////////////////////////////////////////////MAIN///////////////////////////////////////  
    
     public static void main(String[] args)
    {
       
        int t = scanner.nextInt();
      
    while(t>0)
    { 
       solve();
     
        t-=1;
    }
        
    }
    ////////////////////////////
    
   
  
   //////////////////////////////////////////////////////////////Solve//////////////////////////////////////// 
    static void solve(){
       
      int n=scanner.nextInt();
      int k=scanner.nextInt();
     int prefixNumber=2*k-n-1;

     for(int i=1;i<=prefixNumber;i++)
       System.out.print(i+" ");

     for(int i=k;i>prefixNumber;i--)
      System.out.print(i+" ");
      
      System.out.println("");

    }
 }