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: /*Everything is Hard 
   * Before Easy 
   * Jai Mata Dii 
   */ 
   
  import java.util.*;
  import java.io.*; 
    
  public class Main {
  	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();}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; }} 
  	static long mod = (long)(1e9+7); 
  	// static long mod = 998244353; 
  //	 static Scanner sc = new Scanner(System.in); 
  	static FastReader sc = new FastReader(); 
  	static PrintWriter out = new PrintWriter(System.out);
  	public static void main (String[] args) { 
  		int ttt = 1;
  		ttt = sc.nextInt();
  		z :for(int tc=1;tc<=ttt;tc++){
  			long l = sc.nextLong();
  			long r = sc.nextLong();
  			if(l == r) {
  				out.write(0+"\n");
  				continue;
  			}
  			long find = 0;
  			long x = r, y = l;
  			while(l<=r) {
  				long mid = (l+r)/2;
  				if(2L*mid-1L<=x) {
  //					System.out.println("hii");
  					find = mid;
  					l = mid+1;
  				}
  				else {
  					r = mid-1;
  				}
  			}
  			if(find == 0) {
  				find = x%y+1;
  			}
  			out.write((find-1)+"\n");
  		}
  		out.close();
  	}
  	static long pow(long a, long b){long ret = 1;while(b>0){if(b%2 == 0){a = (a*a)%mod;b /= 2;}else{ret = (ret*a)%mod;b--;}}return ret%mod;}
  	static long gcd(long a,long b){if(b==0) return  a; return gcd(b,a%b); } 
  	private static void sort(double[] a) {List<Double> k = new ArrayList<>();for(double val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  	private static void ini(List<Integer>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void init(List<int[]>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void sort(long[] a) {List<Long> k = new ArrayList<>();for(long val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  }