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.*;
public class Main {

    private static Scanner sc = new Scanner(System.in);
    private static int N = 200000+69;
    private static long MOD = 1000000007;
    private static long mod = 998244353;
    private static long inf = 1000000000000000000L;
    public static void main(String[] args)
    {
        int t=1;
        t = sc.nextInt();
        while(t-- > 0)
        {
            solve();
        }
    }
    private static void solve()
    {
         int n = sc.nextInt(),k = sc.nextInt();
         int a[] = new int[n+1],ele = 2*k-n,p[] = new int[k+1],b[] = new int[n+1];
         for(int i=1;i<=n;i++)
         {
             if(i<=k) a[i] = a[i-1]+1;
             else a[i] = a[i-1]-1;
         }
         for(int i=1;i<=n;i++)
         {
             if(a[i]>=ele)
                 b[i] = ele+k-a[i];
             else b[i] = a[i];
         }
         for(int i=1;i<=n;i++) p[a[i]] = b[i];
         for(int i=1;i<=k;i++)
         {
             System.out.printf(p[i] + " ");
         }
        System.out.printf("\n");
    }
}