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.util.*;
  import java.io.*;
  
  public class Main {
      // For fast input output
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              try {
                  br = new BufferedReader(new FileReader("input.txt"));
                  PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
                  System.setOut(out);
              } catch (Exception e) {
                  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;
          }
          int [] readintarray(int n) {
              int res [] = new int [n];
              for(int i = 0; i<n; i++)res[i] = nextInt();
              return res;
          }
          long [] readlongarray(int n) {
              long res [] = new long [n];
              for(int i = 0; i<n; i++)res[i] = nextLong();
              return res;
          }
      }
      static  int gcd(int a,int b)
      {
          if(b==0)
          {
              return a;
          }
          return gcd(b,a%b);
      }
   
      static int lcm(int a, int b)
      {
          return (a / gcd(a, b)) * b;
      }
   
   
      static int abs(int a)
      {
          if(a<0)
              return -1*a;
          return a;
      }
      public static int lower(int arr[],int key){
          int low = 0;
          int high = arr.length-1;
          while(low < high){
              int mid = low + (high - low)/2;
              if(arr[mid] >= key){
                  high = mid;
              }
              else{
                  low = mid+1;
              }
          }
          return low;
      }
      public static int upper(int arr[],int key){
          int low = 0;
          int high = arr.length-1;
          while(low < high){
              int mid = low + (high - low+1)/2;
              if(arr[mid] <= key){
                  low = mid;
              }
              else{
                  high = mid-1;
              }
          }
          return low;
      }
  
      // end of fast i/o code
      public static void main(String[] args) {
          FastReader reader = new FastReader();
           StringBuilder ans = new StringBuilder();   
          int t = reader.nextInt();
          while (t-- > 0) {
              int l = reader.nextInt();
              int r = reader.nextInt();
  
              if((r&1) == 1){
                  if(r/2 < l){
                      ans.append(r%l + "\n");
                  } else 
                  ans.append(r/2 + "\n");
              
              } else {
                  if(r/2 < l)
                   ans.append(r%l + "\n");
                   else 
                   ans.append(r/2-1 + "\n");
              }
  
              
          }
          System.out.println(ans);
      }
      
    
  }