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.*;
import java.util.*;
public class CodeForcesTest {
	public static void main(String[] args) throws IOException {
        FastReader sc=new FastReader();
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(new FileOutputStream(java.io.FileDescriptor.out), "ASCII"), 512);
		int t=sc.nextInt();
        while(t--!=0){
			int n=sc.nextInt(),k=sc.nextInt();
			for(int i=1;i<k-(n-k);i++){
				System.out.print(i+" ");
			}
			for(int i=k;i>=k-(n-k);i--){
				System.out.print(i+" ");
			}
			System.out.println();
        }
        out.close();
	}
    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; 
		} 
	} 
	static int Mex(int[] a){
        Arrays.sort(a);
        if(a[0]!=1){
            return 1;
        }
        for(int i=0;i<a.length-1;i++){
            if(a[i]!=a[i+1] && a[i+1]>a[i]+1){
                return a[i]+1;
            }
        }
        return a[a.length-1]+1;
    }
	static int gcd(Integer a,Integer b)
    {
        if(b==0)
            return a;
        else
            return gcd(b,a%b);
    }
	static int[] readArray(int n){
		FastReader sc = new FastReader();
		int[] a = new int[n];
		for(int i=0;i<n;i++){
			a[i]=sc.nextInt();
		}
		return a;
	}
}