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.*;
  public class Main{
      public static int maxRemainder(int l, int r){
          if(l*2>r){
      		return r%l;
      	}
      	else if(l*2==r){
      		return (r-1)%l;
      	}
      	int low=l;
      	int high=r;
      	int mid=0;
      	int maxAns=Integer.MIN_VALUE;
      	while(low<=high){
      		mid=(low+high)/2;
      		if(mid*2 <r){
      			maxAns=Math.max(maxAns,mid);
      			low=mid+1;
      		}
      		else{
      			high=mid-1;
      		}
      	}
      	if(maxAns>0) return maxAns;
      	else {
      		return r%l;
      	}
      }
      public static void main(String[] args){
          Scanner s=new Scanner(System.in);
          int t=s.nextInt();
          for(int i=0;i<t;i++){
          	int l=s.nextInt();
          	int r=s.nextInt();
          	System.out.println(maxRemainder(l,r));
          }
      }
  }