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:
//WHEN IN DOUBT , USE BRUTE FORCE !!!!!!!!!
import java.lang.String;
import java.io.*;
import java.util.*;
import java.util.StringTokenizer;
//class Main    //AtCoder
//class Solution // Codechef
public class Solution2 //Codeforces
{
    public static void main(String args[])
    {
        try {
            FastReader sc = new FastReader();
            int TT = sc.nextInt();
            for(int hehe=1 ; hehe  <= TT ; hehe++){
                int N=sc.nextInt();
                int K=sc.nextInt();
                long tA=((N-K)*(N-K+1))/2;
                long sum=0;
                int z=N-K;
                for(int i=1;i<=K-(N-K)-1;i++) {
                    sop(i + " ");
                }
                for(int jj=K;jj>=K-(N-K) && jj>0;jj--){
                    sop(jj+" ");
                }
                sopln("");
            }
            out.flush();
        }catch (Exception e) {
            sopln(e.toString());
        }
    }
    public final static int d = 256;
    static int MOD = 1000000007;
    static final double PI = Math.PI;
    private static BufferedReader in = new BufferedReader (new InputStreamReader (System.in));
    private static PrintWriter out = new PrintWriter (new OutputStreamWriter (System.out));
    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());
        }

        char nextChar()
        {
            try {
                return (char) (br.read());
            }catch (IOException e){
                return '~';
            }
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
    static void sop(Object o){System.out.print(o);}
    static double ceil(double d){
        return Math.ceil(d);
    }
    static double floor(double d){
        return Math.floor(d);
    }
    static double round(double d){
        return Math.round(d);
    }
    static void sopln(Object o){System.out.println(o);}
    static void printArray(char arr[]){
        for(char L:arr)
            sop(L+" ");
        sopln("");
    }

}