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.*;

public class C{

	public static void main(String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		int t = Integer.parseInt(br.readLine());
		StringBuilder sb = new StringBuilder();
		for(int numCases = 1; numCases <= t; numCases++){
			StringTokenizer st = new StringTokenizer(br.readLine());
			int n = Integer.parseInt(st.nextToken());
			int k = Integer.parseInt(st.nextToken());
			int[] rev = new int[k - (k - (n - k)) + 1];
			int index = 0;
			// System.out.println("rev.length: " + rev.length);
			for(int i = k - (n - k); i <= k; i++){
				rev[index++] = i;
			}

			for(int i = 0; i < rev.length / 2; i++){
				int tmp = rev[i];
				rev[i] = rev[rev.length - 1 - i];
				rev[rev.length - 1 - i] = tmp;
			}

			int[] out = new int[n];
			for(int i = 1; i <= k - (n - k) - 1; i++){
				out[i - 1] = i;
			}
			index = k - (n - k) - 1;
			for(int i = k - (n - k); i < k; i++){
				out[index++] = rev[i - (k - (n - k))];
			}
			out[index++] = rev[rev.length - 1];
			for(int i = k - (n - k); i < k; i++){
				out[index++] = rev[rev.length - 1 - (i - (k - (n - k)))];
			}

			for(int i = 0; i < k; i++){
				sb.append(out[i] + " ");
			}
			sb.append("\n");
			// System.out.println(Arrays.toString(rev));
		}
		System.out.print(sb);
		
		br.close();
	}

}