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.Scanner; 
import java.util.StringTokenizer;
import java.util.*;
public class Main {
    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; 
        } 
    }
    public static void main(String[] args) {
       FastReader sc=new FastReader();
      int t=sc.nextInt();
      while(t-->0){
        int n=sc.nextInt();
        int k=sc.nextInt();
        if(n==k){for(int i=0;i<n;i++){System.out.print((i+1)+" ");}}
        else{
        int ar[]=new int[k];
        for(int i=0;i<k;i++){ar[i]=i+1;}
        for(int i=0;i<(2*k)-(n+1);i++){System.out.print((i+1)+" ");}
        for(int i=k-1;i>=(2*k)-(n+1);i--){System.out.print((i+1)+" ");}
        /*for(int i=0;i<k-1;i++){
            if(i==(k-(n-k)-1)){System.out.print(k+" ");}
            
            
            System.out.print(ar[i]+" ");}*/
        }
        System.out.println();
        }
    }
}