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.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Scanner;
import java.util.Stack;

public class CF {
	
	private static Scanner sc = new Scanner(System.in);
	
	public static void main(String[] args){
		int t = sc.nextInt();
		StringBuilder ret = new StringBuilder();
		while(t-- > 0) {
			int n = sc.nextInt();
			int k = sc.nextInt();
			int[] a = new int[k];
			for(int i = 0; i < k; i++) a[i] = i+1;
			int len = 1 + (n-k);
			int start = k-len, end = k-1;
			for(int i = 0; i < len/2; i++) {
				int l = start+i;
				int r = end-i;
				int temp = a[l];
				a[l] = a[r];
				a[r] = temp;
			}
			for(int aa:a) ret.append(aa + " ");
			ret.append("\n");
		}
		System.out.println(ret);
	}
	
	private static int[] input(int n) {
		sc.nextLine();
		String in = sc.nextLine();
		String[] temp = in.trim().split(" ");
		int[] a = new int[n];
		for(int i = 0; i < n; i++) a[i] = Integer.valueOf(temp[i]);
		return a;
	}
	
	private static long sum(int[] a) {
		long sum = 0;
		for(int aa:a) sum += aa;
		return sum;
	}

}