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.*;

public class CF
{
	public static void main(String[] args) {
		new A().run();
	}
	
	static class A{
		FastReader in;
		public void run() {
			in = new FastReader();
			int T = in.nextInt();
			StringBuffer sb = new StringBuffer();

			while(T-- > 0) {
				int n = in.nextInt(), k = in.nextInt();
				
				
				for(int i=0; i<(k - (n-k) - 1); i++) {
					sb.append((i+1) + " ");
				}
				for(int i=0; i<(n-k+1); i++) {
					sb.append((k-i) + " ");
				}
				
				sb.append("\n");
			}
			System.out.println(sb.toString());
		}
		
	}
	
	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;
		}
    	}
}