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.BufferedOutputStream;
import java.io.PrintStream;
import java.util.Scanner;


public class Solution {

    private static boolean TESTS = true;

    private final Scanner in;
    private final PrintStream out;

    public Solution(final Scanner in, final PrintStream out) {
        this.in = in;
        this.out = out;
    }

    public void solve() {
        for (int tests = TESTS ? in.nextInt() : 1; tests > 0; tests--) {
            final int n = in.nextInt();
            final int k = in.nextInt();
            final int[] p = new int[k];

            for (int i = 0; i < 2 * k - n - 1; i++) {
                p[i] = i + 1;
            }

            for (int i = 2 * k - n - 1, j = k; i < k; i++) {
                p[i] = j--;
            }

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

            out.println();
        }
    }

    public static void main(final String[] args) {
        final Scanner in = new Scanner(System.in);
        final PrintStream out = new PrintStream(new BufferedOutputStream(System.out));

        try {
            new Solution(in, out).solve();
        } finally {
            out.flush();
        }
    }

}