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 C
{
	public static class FastIO
	{
		BufferedReader br;
		BufferedWriter bw, be;
		StringTokenizer st;

		public FastIO()
		{
			br = new BufferedReader(new InputStreamReader(System.in));
			bw = new BufferedWriter(new OutputStreamWriter(System.out));
			be = new BufferedWriter(new OutputStreamWriter(System.err));
			st = new StringTokenizer("");
		}

		private void read() throws IOException
		{
			st = new StringTokenizer(br.readLine());
		}

		public String ns() throws IOException
		{
			while(!st.hasMoreTokens())
				read();
			return st.nextToken();
		}

		public int ni() throws IOException
		{
			return Integer.parseInt(ns());
		}

		public long nl() throws IOException
		{
			return Long.parseLong(ns());
		}

		public float nf() throws IOException
		{
			return Float.parseFloat(ns());
		}

		public double nd() throws IOException
		{
			return Double.parseDouble(ns());
		}

		public char nc() throws IOException
		{
			return ns().charAt(0);
		}

		public int[] nia(int n) throws IOException
		{
			int[] a = new int[n];
			for(int i = 0; i < n; i++)
				a[i] = ni();

			return a;
		}

		public long[] nla(int n) throws IOException
		{
			long[] a = new long[n];
			for(int i = 0; i < n; i++)
				a[i] = nl();

			return a;
		}

		public char[] nca() throws IOException
		{
			return f.ns().toCharArray();
		}

		public void out(String s) throws IOException
		{
			bw.write(s);
		}

		public void flush() throws IOException
		{
			bw.flush();
			be.flush();
		}

		public void err(String s) throws IOException
		{
			be.write(s);
		}
	}

	static FastIO f;

	public static void main(String args[]) throws IOException
	{
		f = new FastIO();

		int t, n, k, i, j;
		
		t = f.ni();
		
		while(t-->0)
		{
			n = f.ni();
			k = f.ni();
			n = k - (n-k);

			for(i = 1; i < n; i++)
				f.out(i + " ");
			for(j = k; i <= k; i++, j--)
				f.out(j + " ");
			f.out("\n");
		}

		f.flush();
	}
}