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:
/*
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 * and open the template in the editor.
 */

import java.util.Scanner;
import java.util.StringTokenizer;

/**
 *
 * @author is2ac
 */
public class NoMoreInversions {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t = Integer.parseInt(scan.nextLine());
        for (int tc = 0; tc < t; tc++) {
            StringTokenizer st = new StringTokenizer(scan.nextLine());
            int n = Integer.parseInt(st.nextToken());
            int k = Integer.parseInt(st.nextToken());
            int[] p = new int[k];
            int d = n-k;
            for (int i = 0; i < p.length; i++) {
                p[i] = i + 1;
            }
            if (k!=n) {
                p[k-1] = k-d;
                p[k-d-1] = k;
            }
            if (k!=n) {
                for (int i = k-d-1; i < p.length; i++) {
                    p[i]=k;
                    k--;
                }
            }
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < p.length; i++) {
                sb.append(p[i]);
                sb.append(" ");
            }
            System.out.println(sb.toString());
        }
        //[1,2,3,2,1] -> 2
        //[3,2,1,2,3] -> 
        
        //[1,2,3,4,3] -> 1
        //[1,4,3,2,3] -> 3
        
        //[1,2,3,4,3] -> 1
        //[1,2,4,3,4] -> 1
        
        //[1,2,3,4,3,2] -> 4
        //[1,4,3,2,3,4] -> 4
        
        //[1,2,3,4,5,4,3] -> 4
        //[1,2,5,4,3,4,5] -> 4
        
        //[1,2,3,4,5,4,3,2] -> 9
        //[1,5,4,3,2,3,4,5] -> 9
        //[1,5,4,3,2]
        
        //[1,2,3,4,3,2,1] -> 9
        //[4,2,3,1,3,2,4] -> 9
        //[4,3,2,1,2,3,4] -> 5 + 3 + 1
        
    }
}