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.util.regex.Matcher;
    import java.util.regex.Pattern;
    import java.util.HashMap;




public class Main
{
    static long max = Long.MIN_VALUE;
    static void reverse(int[] a,int l,int r)
    {
        int mid = (r-l)>>1;
        for(int i=0;i<=mid;i++)
        {
            int temp = a[l+i];
            a[l+i] = a[r-i];
            a[r-i] = temp;
        }
//        Stack<Integer> stack = new Stack<Integer>();
//        for(int i=l;i<r;i++)stack.push(a[i]);
//        for(int i=l;i<r;i++)a[i] = stack.pop();
    }

    public static void main(String[] args)
    {
        Scanner scanner = new Scanner(System.in);
        int T = scanner.nextInt();
        scanner.nextLine();
        while(T-->0)
        {
            int n = scanner.nextInt();
            int k = scanner.nextInt();
            int[] ans = new int[k];
            for(int i=0;i<k;i++)ans[i] = i+1;
            reverse(ans,2*k-1-n,k-1);
            for(int i=0;i<k;i++)System.out.print(ans[i]+(i==k-1?"":" "));
            System.out.println();
        }
    }
}