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.*;

public class C {


	public static void main(String[] args)throws IOException {

		InputReader in = new InputReader();
		PrintWriter pw = new PrintWriter(System.out);
		int t = in.nextInt();
		while (t-- > 0) {
			int n = in.nextInt();
			int k = in.nextInt();
			for (int i = 1; i <= 2 * k - n - 1; i++)
				pw.print(i + " ");
			for (int i = k; i > 2 * k - n - 1; i--)
				pw.print(i + " ");
			pw.println();
		}

		pw.close();
	}

	static final Random random = new Random();

	static void ruffleSort(int[] a) {
		int n = a.length;
		for (int i = 0; i < n; i++) {
			int oi = random.nextInt(n), temp = a[oi];
			a[oi] = a[i];
			a[i] = temp;
		}
		ArrayList<Integer> lst = new ArrayList<>();
		for (int i : a)
			lst.add(i);
		Collections.sort(lst);
		for (int i = 0; i < n; i++)
			a[i] = lst.get(i);
	}

	static class InputReader {
		BufferedReader br;
		StringTokenizer st;

		public InputReader() {
			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());
		}
		int[] readArray(int n) {
			int []a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}
		long[] readLongArray(int n) {
			long []a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}
		int [][] read2dArray(int n, int m) {
			int [][]a = new int[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					a[i][j] = nextInt();
			return a;
		}
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}