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.util.*;
   
  public class A {
  	
  public static void main(String[] args) {
  		
  		FastScanner sc = new FastScanner();
  		int t = sc.nextInt();
  		for(int tt = 0 ; tt < t; tt++) {
  			int l = sc.nextInt(), r = sc.nextInt();
  			int mid = (r+1)/2;
  			if(mid >= l) {
  				System.out.println(mid - 1);
  			}
  			else {
  				System.out.println(r%l);
  			}
  		}
  	}
  	
  	public static void sort(int[] a) {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  
    
      public static long gcd(long a, long b) {
          return b == 0 ? a : gcd(b, a % b);
      }
   
      public static int gcd(int a, int b) {
          return b == 0 ? a : gcd(b, a % b);
      }
      public static long pow(long base, long exp, long MOD) {
          base %= MOD;
          long ret = 1;
          while (exp > 0) {
              if ((exp & 1) == 1) ret = ret * base % MOD;
              base = base * base % MOD;
              exp >>= 1;
          }
          return ret;
      }
   
   
      public static long max(long... ar) {
          long ret = ar[0];
          for (long itr : ar) ret = Math.max(ret, itr);
          return ret;
      }
   
      public static int max(int... ar) {
          int ret = ar[0];
          for (int itr : ar) ret = Math.max(ret, itr);
          return ret;
      }
   
      public static long min(long... ar) {
          long ret = ar[0];
          for (long itr : ar) ret = Math.min(ret, itr);
          return ret;
      }
   
      public static int min(int... ar) {
          int ret = ar[0];
          for (int itr : ar) ret = Math.min(ret, itr);
          return ret;
      }
   
   
      public static long sum(long... ar) {
          long sum = 0;
          for (long itr : ar) sum += itr;
          return sum;
      }
   
      public static long sum(int... ar) {
          long sum = 0;
          for (int itr : ar) sum += itr;
          return sum;
      }
  	
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		
  		public String next() {
  			while (!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());
  		}
  		
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  	}
   
  }
  
  
   
   
   
   
   
   
   
   
  