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 test {
  
      static long mod = 1000000007;
      static long inf = Long.MAX_VALUE;
  
      public static void solve(long l, long r) {
          long max = r % l;
          long rhalf = r / 2 + 1;
          if(rhalf >= l) {
              max = Math.max(max, r % rhalf);
          }
          long ldouble = l * 2 -1;
          if(ldouble <= r) {
              max = Math.max(max, ldouble % l);
          }
          System.out.println(max);
      }
      
  /*
  input:
  4
  1 1
  999999999 1000000000
  8 26
  1 999999999
  
  
  output:
  0
  1
  12
  499999999
  
  
  */
  
      public static void main(String[] args) {
          FastReader sc = new FastReader();
          int t = sc.nextInt();
          while(t-- > 0) {
              long l = sc.nextLong();
              long r = sc.nextLong();
              // solve
              solve(l, r);
          }
      }
  
       
  	public static void sort(long[] a) {
  		ArrayList<Long> l=new ArrayList<>();
  		for (long i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  
      public static int maxIndex(long[] arr) {
          long max = Long.MIN_VALUE;
          int index = 0;
          int n = arr.length;
          for(int i = 0; i < n; i++) {
              if(arr[i] > max) {
                  index = i;
                  max = arr[i];
              }
          }
          return index;
      }
  
      public static int minIndex(long[] arr) {
          long min = Long.MAX_VALUE;
          int index = 0;
          int n = arr.length;
          for(int i = 0; i < n; i++) {
              if(arr[i] < min) {
                  index = i;
                  min = arr[i];
              }
          }
          return index;
      }
  
      public static void debugArray(long[] arr) {
          for(int i = 0; i < arr.length; i++) {
              System.out.print(arr[i] + " ");
          }
          System.out.println();
      } 
  
      public static void debugArrayList(ArrayList<Long> arr) {
          for(Long i : arr) {
              System.out.print(i + " ");
          }
          System.out.println();
      }
  
      public static long gcd(long a, long b) {
          if(b == 0) {
              return a;
          }
          else {
              return gcd(b, a % b);
          }
      }
   
      public static long LCM(long a, long b) {
          return Math.abs(a*b) / gcd(a, b);
      }
  
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
      
          public FastReader() {
              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();
          }
  
          String nextLine() {
              String str="";
              try {
                  str=br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          long[] readArray(int n) {
              long[] a = new long[n];
              for(int i = 0; i < n ; i++) {
                  a[i] = nextLong();
              }
              return a;
          }
      }
  }