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.*;
import java.math.*;
public class Main {       /* Author : EIN     Date : 14-01-2021     */

    static FastReader in;
    static PrintWriter o;

    public static void solve() {
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            int k = in.nextInt();
            for (int i = 1; i < 2 * k - n; i++) {
                o.print(i + " ");
            }
            for (int i = k; i >= 2 * k - n; i--) {
                o.print(i + " ");
            }
            o.println();
        }
        o.close();
        return;
    }

    static int getSum(int[] BITree, int index)
    {
        int sum = 0; // Initialize result

        // Traverse ancestors of BITree[index]
        while (index > 0)
        {
            // Add current element of BITree to sum
            sum += BITree[index];

            // Move index to parent node in getSum View
            index -= index & (-index);
        }
        return sum;
    }

    // Updates a node in Binary Index
// Tree (BITree) at given index
// in BITree. The given value 'val'
// is added to BITree[i] and all
// of its ancestors in tree.
    static void updateBIT(int[] BITree, int n,
                          int index, int val)
    {
        // Traverse all ancestors and add 'val'
        while (index <= n)
        {
            // Add 'val' to current node of BI Tree
            BITree[index] += val;

            // Update index to that of parent in update View
            index += index & (-index);
        }
    }

    // Returns inversion count arr[0..n-1]
    static int getInvCount(int[] arr, int n)
    {
        int invcount = 0; // Initialize result

        // Find maximum element in arr[]
        int maxElement = 0;
        for (int i = 0; i < n; i++)
            if (maxElement < arr[i])
                maxElement = arr[i];

        // Create a BIT with size equal to
        // maxElement+1 (Extra one is used so
        // that elements can be directly be
        // used as index)
        int[] BIT = new int[maxElement + 1];
        for (int i = 1; i <= maxElement; i++)
            BIT[i] = 0;

        // Traverse all elements from right.
        for (int i = n - 1; i >= 0; i--)
        {
            // Get count of elements smaller than arr[i]
            invcount += getSum(BIT, arr[i] - 1);

            // Add current element to BIT
            updateBIT(BIT, maxElement, arr[i], 1);
        }

        return invcount;
    }

    public static void main(String[] args) {
        in = new FastReader();
        o = new PrintWriter(System.out);
        solve();
        return;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        long[] readLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}