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.*; 
import java.lang.Math; 
public class Main{
     public static void main(String []args){
        Scanner in=new Scanner(System.in);
        int t,n,k;
        t=in.nextInt();
        for(int i=0;i<t;i++)
        {
            n=in.nextInt();
            k=in.nextInt();
            int ni=n-k,temp=1;
            int a[]=new int[n];
            int p[]=new int[k];
            int b[]=new int[n];
            for(int j=1;j<=n;j++)
            {
                if(j<=k)
                {
                    a[j-1]=j;
                }
                else
                {
                    a[j-1]=k-temp;
                    temp++;
                }
                b[j-1]=a[j-1];
                //System.out.print(a[j-1]);
            }
            //System.out.println("n-(2*ni)-1 = "+(n-(2*ni)-1));
            temp=0;
            int temp1=2;
            for(int j=n-(2*ni)-1;j<n;j++)
            {
                if(j<k)
                {
                    b[j]=a[k-temp-1];
                    temp++;
                }
                else
                {
                    b[j]=b[j-temp1];
                    //System.out.println("b[j-temp1]= "+b[j-temp1]+" "+j+" "+temp1);
                    temp1+=2;
                }
                //System.out.print("b["+j+"]="+b[j]+", ");
            }
            
            //System.out.println();
           /* for(int j=0;j<n;j++)
            {
                System.out.print(b[j]);
            }
            System.out.println();*/
            
            for(int j=0;j<n;j++)
            {
                p[a[j]-1]=b[j];
            }
            for(int j=0;j<k;j++)
            {
                System.out.print(p[j]+" ");
            }
            System.out.println();
        }
     }
}