description: You are given two integers l and r, l≤ r. Find the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.
  
  As a reminder, a mod b is a remainder we get when dividing a by b. For example, 26 mod 8 = 2.
  
  Input
  
  Each test contains multiple test cases.
  
  The first line contains one positive integer t (1≤ t≤ 10^4), denoting the number of test cases. Description of the test cases follows.
  
  The only line of each test case contains two integers l, r (1≤ l ≤ r ≤ 10^9).
  
  Output
  
  For every test case, output the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.
  
  Example
  
  Input
  
  
  4
  1 1
  999999999 1000000000
  8 26
  1 999999999
  
  
  Output
  
  
  0
  1
  12
  499999999
  
  Note
  
  In the first test case, the only allowed pair is (a, b) = (1, 1), for which a mod b = 1 mod 1 = 0.
  
  In the second test case, the optimal choice is pair (a, b) = (1000000000, 999999999), for which a mod b = 1.
solution: /********************************************************************
  *******  ######      #       #   #  #####    ###   ******************
  *******  #          #  #     #  #     #      #  #  ******************
  *******  ######    ######    ###      #      ###   ******************
  *******       #   #      #   #  #     #      #  #  ******************
  *******  ######  #        #  #   #  #####    ###   ******************
  *********************************************************************
  ********************************************************************/
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.OutputStreamWriter;
  import java.io.PrintWriter;
  import java.util.*;
  import java.lang.*;
  
  
  
  public class test{
  
    static BufferedReader br;
    static PrintWriter out;
    static StringTokenizer st;
  
    public static void main(String[] args) throws IOException 
    {
      br = new BufferedReader(new InputStreamReader(System.in));
      out = new PrintWriter(new OutputStreamWriter(System.out));
      // br = new BufferedReader(new FileReader("in.txt"));
      // out = new PrintWriter(new FileWriter("out.txt"));
  	try{
  			//Your Solve
  		    int t = readInt();
  		    while(t-->0){
  		    int f=0,d=0;
  		    	int l = readInt();
  		    	int r = readInt();
  		    	d = r/2;
  		    	d++;
  		    	if(l==r){
  		    	System.out.println(0);
  		    	}
  		    	else if(l>=d)
  		    	System.out.println(r%l);
  		    	else	
  		    	System.out.println(r%d);
  		    }
  		}catch(Exception e){
  			return;
  		}
      
    }
  
    static String next() throws IOException 
    {
      while (st == null || !st.hasMoreTokens())
        st = new StringTokenizer(br.readLine().trim());
      return st.nextToken();
    }
      //extended
      static int gcd(int a, int b, int x, int y)
      {
          // Base Case
          if (a == 0)
          {
              x = 0;
              y = 1;
              return b;
          }
   
          int x1=1, y1=1; // To store results of recursive call
          int gcd = gcd(b%a, a, x1, y1);
   
          // Update x and y using results of recursive
          // call
          x = y1 - (b/a) * x1;
          y = x1;
   
          return gcd;
      }
      
    static long readLong() throws IOException 
    {
      return Long.parseLong(next());
    }
  
    static int readInt() throws IOException
    {
      return Integer.parseInt(next());
    }
  
    static double readDouble() throws IOException 
    {
      return Double.parseDouble(next());
    }
  
    static char readCharacter() throws IOException 
    {
      return next().charAt(0);
    }
  
    static String readLine() throws IOException 
    {
      return br.readLine().trim();
    }
  
  }