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: // package Codeforce.cf741;
  
  import java.io.PrintWriter;
  import java.util.*;
  
  public class A {
      //    MUST SEE BEFORE SUBMISSION
  //    check whether int part would overflow or not, especially when it is a * b!!!!
  //    check if top down dp would cause overflow or not !!!!!!!!!!!!!!!!!!!!!!
  //    ------------consider only if you running into mistake--------
  //    consider all the edge cases such as k > 0 or k >= 0
  //    consider some edge case such as extreme situation
  //    when you swap forward, think more about the stop condition！！！！！！
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
          int t = sc.nextInt();
  //        int t = 1;
          for (int i = 0; i < t; i++) {
              solve(sc, pw);
          }
          pw.close();
      }
  
      static void solve(Scanner in, PrintWriter out){
          int l = in.nextInt(), r = in.nextInt();
          if ((r - (r - 1) / 2) < l){
              out.println((r % l));
          }else{
              out.println((r - 1) / 2);
          }
      }
      static int mod(int l, int r){
          int max = 0;
          for (int i = l; i <= r; i++) {
              for(int j = i; j <= r; j++){
                  max = Math.max(max, j % i);
              }
          }
          return max;
      }
      static int md(int l, int r){
          if ((r - (r - 1) / 2) < l){
              return (r % l);
          }else{
              return (r - 1) / 2;
          }
      }
  }
  