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.math.*;
import java.util.*;

// author : Multi-Thread

public class A {

	public static void main(String[] args) {
		int test = fs.nextInt();
//		int test = 1;
		for (int cases = 0; cases < test; cases++) {
			int n = fs.nextInt(), k = fs.nextInt();
//			int ar[]=new int[n];
			if (n == 1) {
				System.out.println(1);
			} else {
				int diff = k - (n - k);
				diff = k - diff + 1;
				int ar[] = new int[k];
				for (int i = 0; i < k; i++) {
					ar[i] = i + 1;
				}
				int start = (k - 1) - (diff - 1);
				int end = k - 1;
//				System.out.println(start + " " + end);
				while (start <= end) {
					int t = ar[start];
					ar[start] = ar[end];
					ar[end] = t;
					++start;
					--end;
				}
				for (int i : ar) {
					System.out.print(i + " ");
				}
				System.out.println();
			}
		}
	}

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

	static int lcm(int a, int b) {
		return (a / gcd(a, b)) * b;
	}

	static void intSort(int[] a, boolean reverse) {
		ArrayList<Integer> al = new ArrayList<Integer>();
		for (int i : a)
			al.add(i);
		Collections.sort(al);
		if (reverse) {
			for (int i = 0; i < a.length; i++)
				a[i] = al.get(a.length - i - 1);
		} else {
			for (int i = 0; i < a.length; i++)
				a[i] = al.get(i);
		}
	}

	static void longSort(long[] a, boolean reverse) {
		ArrayList<Long> al = new ArrayList<>();
		for (long i : al)
			al.add(i);
		Collections.sort(al);
		if (reverse) {
			for (int i = 0; i < a.length; i++)
				a[i] = al.get(a.length - i - 1);
		} else {
			for (int i = 0; i < a.length; i++)
				a[i] = al.get(i);
		}
	}

	static void sieve() {
		boolean prime[] = new boolean[101];
		Arrays.fill(prime, true);
		prime[1] = false;
		for (int i = 2; i * i <= prime.length - 1; i++) {
			for (int j = i * i; j <= prime.length - 1; j += i) {
				prime[j] = false;
			}
		}
	}

	public static int[] radixSort(int[] f) {
		return radixSort(f, f.length);
	}

	public static int[] radixSort(int[] f, int n) {
		int[] to = new int[n];
		{
			int[] b = new int[65537];
			for (int i = 0; i < n; i++)
				b[1 + (f[i] & 0xffff)]++;
			for (int i = 1; i <= 65536; i++)
				b[i] += b[i - 1];
			for (int i = 0; i < n; i++)
				to[b[f[i] & 0xffff]++] = f[i];
			int[] d = f;
			f = to;
			to = d;
		}
		{
			int[] b = new int[65537];
			for (int i = 0; i < n; i++)
				b[1 + (f[i] >>> 16)]++;
			for (int i = 1; i <= 65536; i++)
				b[i] += b[i - 1];
			for (int i = 0; i < n; i++)
				to[b[f[i] >>> 16]++] = f[i];
			int[] d = f;
			f = to;
			to = d;
		}
		return f;
	}

	static class Pair {
		int first, second;

		public Pair(int first, int second) {
			this.first = first;
			this.second = second;
		}

		public String toString() {
			return "[" + first + "," + second + "]";
		}
	}

	static class LongPair {
		long first;
		long second;

		LongPair(long a, long b) {
			this.first = a;
			this.second = b;
		}

		public String toString() {
			return "[" + first + "," + second + "]";
		}
	}

	static void primeFactors(int n) {
		while (n % 2 == 0) {
			System.out.print(2 + " ");
			n /= 2;
		}
		for (int i = 3; i <= Math.sqrt(n); i += 2) {
			while (n % i == 0) {
				System.out.print(i + " ");
				n /= i;
			}
		}
		if (n > 2)
			System.out.print(n);
	}

	static boolean isPrime(long n) {
		// Corner cases
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;
		if (n % 2 == 0 || n % 3 == 0)
			return false;

		long sq = (long) Math.sqrt(n);
		for (int i = 5; i <= sq; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;

		return true;
	}

	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());
		}

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

	static class OutputWriter {
		private final PrintWriter writer;

		public OutputWriter(OutputStream outputStream) {
			writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
		}

		public OutputWriter(Writer writer) {
			this.writer = new PrintWriter(writer);
		}

		public void print(Object... objects) {
			for (int i = 0; i < objects.length; i++) {
				if (i != 0)
					writer.print(' ');
				writer.print(objects[i]);
			}
		}

		public void println() {
			writer.print("\n");
		}

		public void printLine(Object... objects) {
			print(objects);
			writer.println();
		}

		public void close() {
			writer.close();
		}

		public void flush() {
			writer.flush();
		}

	}

	private static final FastReader fs = new FastReader();
	private static final OutputWriter op = new OutputWriter(System.out);
}