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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class C {
    public static void main(String[] args) throws IOException {
        Scanner scan = new Scanner(System.in);
//        BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
        int t = scan.nextInt();
//        int t= Integer.parseInt(br.readLine());
        while (t-->0)
        {
//            int n= Integer.parseInt(br.readLine());
            int n= scan.nextInt();
            int k= scan.nextInt();
//            String[] s0= br.readLine().split(" ");
//            int n= Integer.parseInt(s0[0]);
//            int m= Integer.parseInt(s0[1]);
//            Integer[] k= new Integer[n];
//            Integer[] c= new Integer[m];
//            String[] s1= br.readLine().split(" ");
//            String[] s2= br.readLine().split(" ");
//            for (int i=0;i<n;i++){
//                k[i]= Integer.parseInt(s1[i]);
//            }

            Integer[] a= new Integer[k];
            for (int i=0;i<k;i++) a[i]=-1;

            int x= k+k-n; int kk=k;

            for (int i=0;i<kk;i++){
                if (i<x-1) a[i]=i+1;
                else a[i]=k--;

                System.out.print(a[i]+" ");
            }
            System.out.println();
        }
    }

    static long nCr(int n, int r)
    {
        long p = 1, k = 1;
        if (n - r < r) {
            r = n - r;
        }
        if (r != 0) {
            while (r > 0) {
                p *= n;
                k *= r;
                long m = gcd(p, k);
                p /= m;
                k /= m;
                n--;
                r--;
            }
        } else {
            p = 1;
        }
        return p;
    }

    static long gcd(long a, long b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    static int factorial(int n)
    {
        int res[] = new int[500];
        res[0] = 1;
        int res_size = 1;
        for (int x = 2; x <= n; x++)
            res_size = multiply(x, res, res_size);

        for (int i=res_size-1;i>=0;i--)
            return (res[i]);
        return 1;
    }

    static int multiply(int x,int res[],int res_size)
    {
        int carry = 0; // Initialize carry
        for (int i=0;i<res_size;i++)
        {
            int prod = res[i] * x + carry;
            res[i] = prod % 10; // Store last digit of
            carry = prod/10; // Put rest in carry
        }
        while (carry!=0)
        {
            res[res_size] = carry % 10;
            carry = carry / 10;
            res_size++;
        }
        return res_size;
    }
}