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.*;
import java.math.*;
import java.math.BigInteger; 
//import javafx.util.*; 
public final class B
{    
	static StringBuilder ans=new StringBuilder();
	static FastReader in=new FastReader();
	static ArrayList<ArrayList<Integer>> g;	
	static long mod=1000000007;
	static int dp[][];

	static boolean set[];
	public static void main(String args[])throws IOException
	{  
		int T=i();
		while(T-->0)
		{
			int N=i(),K=i();
			int A[]=new int[K];
			for(int i=0; i<K; i++)
			{
				A[i]=(i+1);
			}
			reverse(N-K,A,N);
			print(A);
		}
  
   }
	static void reverse(int d,int A[],int N)
	{
		int K=A.length;
		d=K-d-1;
		int x=K;
		while(d<K)
		{
			A[d++]=x--;
		}
	
		
	}
   static void print(char A[])
   {
	   for(char c:A)System.out.print(c);
		 System.out.println();
   }
	static void print(int A[])
	{
		for(int a:A)System.out.print(a+" ");
		System.out.println();	
	}
	static long lower_Bound(long A[],int low,int high, long x) 
	{ 
		if (low > high) 
			if (x >= A[high]) 
				return A[high]; 

		int mid = (low + high) / 2; 

		if (A[mid] == x) 
			return A[mid]; 

		if (mid > 0 && A[mid - 1] <= x && x < A[mid]) 
			return A[mid - 1]; 

		if (x < A[mid]) 
			return lower_Bound( A, low, mid - 1, x); 

		return lower_Bound(A, mid + 1, high, x); 
	} 
	static void sort(long[] a) //check for long
	{
		ArrayList<Long> l=new ArrayList<>();
		for (long i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	static void setGraph(int N)
	{
		set=new boolean[N+1];

		g=new ArrayList<ArrayList<Integer>>();
		//tg=new ArrayList<ArrayList<Integer>>();
		for(int i=0; i<=N; i++)
		{
			g.add(new ArrayList<Integer>());
			//tg.add(new ArrayList<Integer>());
		}
	}
	static void DFS(int N,int d)
	{
		//  set[N]=true;
		d++;
		for(int i=0; i<g.get(N).size(); i++)
		{
			int c=g.get(N).get(i);

		}
	}
	static  long pow(long a,long b)
	{
		long mod=1000000007;
		long pow=1;
		long x=a;
		while(b!=0)
		{
			if((b&1)!=0)pow=(pow*x)%mod;
			x=(x*x)%mod;
			b/=2;
		}
		return pow;
	}
	static long toggleBits(long x)//one's complement || Toggle bits
	{
		int n=(int)(Math.floor(Math.log(x)/Math.log(2)))+1;

		return ((1<<n)-1)^x;
	}
	static int countBits(long a)
	{
		return (int)(Math.log(a)/Math.log(2)+1);
	}
	static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	static boolean isPrime(long N)
	{
		if (N<=1)  return false; 
		if (N<=3)  return true; 
		if (N%2 == 0 || N%3 == 0) return false; 
		for (int i=5; i*i<=N; i=i+6) 
			if (N%i == 0 || N%(i+2) == 0) 
				return false; 
		return true; 
	}
	static long GCD(long a,long b) 
	{
		if(b==0)
		{
			return a;
		}
		else return GCD(b,a%b );
	}

	static int i()
	{
		return in.nextInt();
	}

	static long l()
	{
		return in.nextLong();
	}

	static int[] input(int N){
		int A[]=new int[N];
		for(int i=0; i<N; i++)
		{
			A[i]=in.nextInt();
		}
		return A;
	}

	static long[] inputLong(int N)     {
		long A[]=new long[N];
		for(int i=0; i<A.length; i++)A[i]=in.nextLong();
		return A;
	}


}

//Code For FastReader
//Code For FastReader
//Code For FastReader
//Code For FastReader
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;
	}

}