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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Random;
import java.util.StringTokenizer;

public class NoMoreInversions {
	
	static final int MAXN = 1000_006;
	
	static final long MOD = (long) 1e9 + 7;
	
	public static void main(String[] args) throws IOException {
		MyScanner s = new MyScanner();
		Print p = new Print();
		int t = s.nextInt();
		
		while (t-- > 0) {
			int n = s.nextInt();
			int k = s.nextInt();
			int cou = n - k;
			LinkedList<Integer> sol = new LinkedList<>();
			sol.add(k);
			while (cou > 0) {
				k--;
				sol.add(k);
				cou--;
			}
			
			while (k > 0) {
				k--;
				if (k == 0)
					break;
				sol.addFirst(k);
			}
			
			for (int num : sol) {
				p.print(num + " ");
			}
			p.println("");
		}
		
		p.close();
	}
	
	public static class Pair implements Comparable<Pair> {
		
		int first;
		
		int second;
		
		public Pair(int a, int b) {
			this.first = a;
			this.second = b;
		}
		
		@Override
		public int hashCode() {
			final int prime = 31;
			int result = 1;
			result = prime * result + first;
			result = prime * result + second;
			return result;
		}
		
		@Override
		public boolean equals(Object obj) {
			if (this == obj)
				return true;
			if (obj == null)
				return false;
			if (getClass() != obj.getClass())
				return false;
			Pair other = (Pair) obj;
			if (first != other.first)
				return false;
			if (second != other.second)
				return false;
			return true;
		}
		
		@Override
		public int compareTo(Pair o) {
			// TODO Auto-generated method stub
			return o.first - first;
		}
		
	}
	
	public static class Helper {
		
		long MOD = (long) 1e9 + 7;
		
		int MAXN = 1000_006;;
		
		Random rnd;
		
		public Helper(long mod, int maxn) {
			MOD = mod;
			MAXN = maxn;
			rnd = new Random();
		}
		
		public Helper() {
			
		}
		
		public static int[] sieve;
		
		public static ArrayList<Integer> primes;
		
		public void setSieve() {
			primes = new ArrayList<>();
			sieve = new int[MAXN];
			int i, j;
			for (i = 2; i < MAXN; ++i)
				if (sieve[i] == 0) {
					primes.add(i);
					for (j = i; j < MAXN; j += i) {
						sieve[j] = i;
					}
				}
		}
		
		public static long[] factorial;
		
		public void setFactorial() {
			factorial = new long[MAXN];
			factorial[0] = 1;
			for (int i = 1; i < MAXN; ++i)
				factorial[i] = factorial[i - 1] * i % MOD;
		}
		
		public long getFactorial(int n) {
			if (factorial == null)
				setFactorial();
			return factorial[n];
		}
		
		public long ncr(int n, int r) {
			if (r > n)
				return 0;
			if (factorial == null)
				setFactorial();
			long numerator = factorial[n];
			long denominator = factorial[r] * factorial[n - r] % MOD;
			return numerator * pow(denominator, MOD - 2, MOD) % MOD;
		}
		
		public long[] getLongArray(int size, MyScanner s) throws Exception {
			long[] ar = new long[size];
			for (int i = 0; i < size; ++i)
				ar[i] = s.nextLong();
			return ar;
		}
		
		public int[] getIntArray(int size, MyScanner s) throws Exception {
			int[] ar = new int[size];
			for (int i = 0; i < size; ++i)
				ar[i] = s.nextInt();
			return ar;
		}
		
		public int[] getIntArray(String s) throws Exception {
			s = s.trim().replaceAll("\\s+", " ");
			String[] strs = s.split(" ");
			int[] arr = new int[strs.length];
			for (int i = 0; i < strs.length; i++) {
				arr[i] = Integer.parseInt(strs[i]);
			}
			return arr;
		}
		
		public long gcd(long a, long b) {
			return b == 0 ? a : gcd(b, a % b);
		}
		
		public int gcd(int a, int b) {
			return b == 0 ? a : gcd(b, a % b);
		}
		
		public long max(long[] ar) {
			long ret = ar[0];
			for (long itr : ar)
				ret = Math.max(ret, itr);
			return ret;
		}
		
		public int max(int[] ar) {
			int ret = ar[0];
			for (int itr : ar)
				ret = Math.max(ret, itr);
			return ret;
		}
		
		public long min(long[] ar) {
			long ret = ar[0];
			for (long itr : ar)
				ret = Math.min(ret, itr);
			return ret;
		}
		
		public int min(int[] ar) {
			int ret = ar[0];
			for (int itr : ar)
				ret = Math.min(ret, itr);
			return ret;
		}
		
		public long sum(long[] ar) {
			long sum = 0;
			for (long itr : ar)
				sum += itr;
			return sum;
		}
		
		public long sum(int[] ar) {
			long sum = 0;
			for (int itr : ar)
				sum += itr;
			return sum;
		}
		
		public long pow(long base, long exp, long MOD) {
			base %= MOD;
			long ret = 1;
			while (exp > 0) {
				if ((exp & 1) == 1)
					ret = ret * base % MOD;
				base = base * base % MOD;
				exp >>= 1;
			}
			return ret;
		}
		
	}
	
	static class Print {
		
		private BufferedWriter bw;
		
		public Print() {
			this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
		}
		
		public void print(Object object) throws IOException {
			bw.append("" + object);
		}
		
		public void println(Object object) throws IOException {
			print(object);
			bw.append("\n");
		}
		
		public void close() throws IOException {
			bw.close();
		}
	}
	
	public static class MyScanner {
		
		BufferedReader br;
		
		StringTokenizer st;
		
		public MyScanner() {
			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;
		}
	}
	
}