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.*;
 
 
public class Main{

	public static Reader sc = new Reader();
	//public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

   public static void main(String[] args) throws IOException{
	
	int t = sc.nextInt();
	while (t-->0){
		int n= sc.nextInt();
		int k= sc.nextInt();
		//Invert the last k+1 elements.
		int endingterm = 2*k-n;
		List<Integer> permutation = new ArrayList<Integer>();
		for (int i = k; i >= endingterm; i--) {
			permutation.add(i);
		}
		int[] a = new int[n];
		for (int i =1; i <= k - permutation.size(); i++) {
			out.print(i + " ");
		}
		for (int i = 0; i < permutation.size(); i++) {
			out.print(permutation.get(i) +" ");
		}
		out.println();
	}
	
	out.close();
   }
   static long ceil(long a, long b) {
		return (a+b-1)/b;
	}
  
   static long powMod(long base, long exp, long mod) {
	   if (base == 0 || base == 1) return base;
	   if (exp == 0) return 1;
	   if (exp == 1) return base % mod;
	   long R = powMod(base, exp/2, mod) % mod;
	   R *= R;
	   R %= mod;
	   if ((exp & 1) == 1) {
		   return base * R % mod;
	   }
	   else return R % mod;
   }
   
   static long pow(long base, long exp) {
	   if (base == 0 || base == 1) return base;
	   if (exp == 0) return 1;
	   if (exp == 1) return base;
	   long R = pow(base, exp/2);
	   if ((exp & 1) == 1) {
		   return R * R * base;
	   }
	   else return R * R;
   }
   

 
   static class Reader 
   { 
       final private int BUFFER_SIZE = 1 << 16; 
       private DataInputStream din; 
       private byte[] buffer; 
       private int bufferPointer, bytesRead; 

       public Reader() 
       { 
           din = new DataInputStream(System.in); 
           buffer = new byte[BUFFER_SIZE]; 
           bufferPointer = bytesRead = 0; 
       } 

       public Reader(String file_name) throws IOException 
       { 
           din = new DataInputStream(new FileInputStream(file_name)); 
           buffer = new byte[BUFFER_SIZE]; 
           bufferPointer = bytesRead = 0; 
       } 

       public String readLine() throws IOException 
       { 
           byte[] buf = new byte[64]; // line length 
           int cnt = 0, c;
           while ((c = read()) != -1) 
           { 
               if (c == '\n') 
               {
            	   break;
               }
               buf[cnt++] = (byte) c; 
           } 
           return new String(buf, 0, Integer.max(cnt-1, 0)); 
       } 

       public int nextInt() throws IOException 
       { 
           int ret = 0; 
           byte c = read(); 
           while (c <= ' ') 
               c = read(); 
           boolean neg = (c == '-'); 
           if (neg) 
               c = read(); 
           do
           { 
               ret = ret * 10 + c - '0'; 
           }  while ((c = read()) >= '0' && c <= '9'); 

           if (neg) 
               return -ret; 
           return ret; 
       } 

       public long nextLong() throws IOException 
       { 
           long ret = 0; 
           byte c = read(); 
           while (c <= ' ') 
               c = read(); 
           boolean neg = (c == '-'); 
           if (neg) 
               c = read(); 
           do { 
               ret = ret * 10 + c - '0'; 
           } 
           while ((c = read()) >= '0' && c <= '9'); 
           if (neg) 
               return -ret; 
           return ret; 
       } 

       public double nextDouble() throws IOException 
       { 
           double ret = 0, div = 1; 
           byte c = read(); 
           while (c <= ' ') 
               c = read(); 
           boolean neg = (c == '-'); 
           if (neg) 
               c = read(); 

           do { 
               ret = ret * 10 + c - '0'; 
           } 
           while ((c = read()) >= '0' && c <= '9'); 

           if (c == '.') 
           { 
               while ((c = read()) >= '0' && c <= '9') 
               { 
                   ret += (c - '0') / (div *= 10); 
               } 
           } 

           if (neg) 
               return -ret; 
           return ret; 
       } 

       private void fillBuffer() throws IOException 
       { 
           bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
           if (bytesRead == -1) 
               buffer[0] = -1; 
       } 

       private byte read() throws IOException 
       { 
           if (bufferPointer == bytesRead) 
               fillBuffer(); 
           return buffer[bufferPointer++]; 
       } 

       public void close() throws IOException 
       { 
           if (din == null) 
               return; 
           din.close(); 
       } 
   }

	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
}