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.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
	try{
	    InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
	    int t=Integer.parseInt(in.next());
	    for(int i=0;i<t;i++)
	    {
	        
	       int n=Integer.parseInt(in.next());
	       int k=Integer.parseInt(in.next());
	       int arr[]=new int[k];
	       for(int j=0;j<k;j++)
	       {
	           arr[j]=j+1;
	       }
	       long diff=n-k;
	       long okk=k-diff;
	       
	       for(int j=0;j<okk-1;j++)
	       {
	           out.print(arr[j]+" ");
	       }
	       for(int j=k-1;j>=okk-1;j--)
	       {
	           out.print(arr[j]+" ");
	       }
	       out.println();
	    }
	   
	   
	    out.close();
	}
	catch(Exception e)
	{
	    return;
	}
	}
}



class FastReader {

    private InputStream stream;
    private byte[] buf = new byte[1024];
    private int curChar;
    private int numChars;
    private FastReader.SpaceCharFilter filter;

    public FastReader(InputStream stream) {
        this.stream = stream;
    }
     public int read() {

        if (numChars == -1) {
            throw new InputMismatchException();
        }

        if (curChar >= numChars) {

            curChar = 0;

            try {
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }

            if (numChars <= 0) {
                return -1;
            }
        }

        return buf[curChar++];
    }
    
    public int nextInt() {

        int c = read();

        while (isSpaceChar(c)) {
            c = read();
        }
        int sgn = 1;

        if (c == '-') {
            sgn = -1;
            c = read();
        }

        int res = 0;

        do {
            if (c < '0' || c > '9') {
                throw new InputMismatchException();
            }

            res *= 10;
            res += c - '0';
            c = read();
        } while (!isSpaceChar(c));

        return res * sgn;
    }
    
     public long nextLong() {

        int c = read();

        while (isSpaceChar(c)) {
            c = read();
        }
        
        int sgn = 1;

        if (c == '-') {
            sgn = -1;
            c = read();
        }
        long res = 0;
         do {
            if (c < '0' || c > '9') {
                throw new InputMismatchException();
            }

            res *= 10;
            res += c - '0';
            c = read();
        } while (!isSpaceChar(c));
        return res * sgn;
    }
    
    public String next() {

        int c = read();

        while (isSpaceChar(c)) {
            c = read();
        }

        StringBuilder res = new StringBuilder();

        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isSpaceChar(c));

        return res.toString();
    }
    public boolean isSpaceChar(int c) {

        if (filter != null) {
            return filter.isSpaceChar(c);
        }

        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

    public interface SpaceCharFilter {

        public boolean isSpaceChar(int ch);

    }
}