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.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class taska {
  	public static void main(String[] args) {
  		FastScanner in = new FastScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int tt = in.nextInt();
  		for (int tc = 0; tc < tt; tc++) {
  			int l = in.nextInt(), r = in.nextInt();
  			int i = l;
  			int low = l, high = r;
  			while (low <= high) {
  				int mid = (low + high) / 2;
  				if (2 * mid - 1 <= r) {
  					i = mid;
  					low = mid+1;
  				} else {
  					high = mid-1;
  				}
  			}
  			int j = Math.min(2*i-1, r);
  			out.println(j % i);
  		}
  		out.flush();
  	}
  
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer tok = new StringTokenizer("");
  		String next() {
  			while(!tok.hasMoreTokens()) {
  				try {
  					tok = new StringTokenizer(br.readLine());
  				} catch (Exception e) {
  					e.printStackTrace();
  				}
  			}
  			return tok.nextToken();
  		}
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		float nextFloat() {
  			return Float.parseFloat(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  }
  