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.Scanner;
import java.util.ArrayList;
import java.util.Arrays;
import java.lang.Math;
import java.util.LinkedList;
 
public class Competition {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int num = Integer.parseInt(scan.nextLine());
        for (int i = 0; i < num; i++) {
            String[] twoNums = scan.nextLine().split(" ");
            solver(Integer.parseInt(twoNums[0]), Integer.parseInt(twoNums[1]));
        }

        scan.close();
    }

    private static void solver(int a, int k) {
        if (k == 1) {
            System.out.println(1);
            return;
        }

        int[] arr = new int[k];
        for (int i = 0; i < k; i++) {
            arr[i] = i + 1;
        }

        int diff = a - k;
        int begin = k - diff;
        for (int i = arr.length - 1; diff >= 0; diff--) {
            arr[i] = begin;
            begin++;
            i--;
        }

        for (int i = 0; i < k; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    }
}