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 sys,os,io
  import math 
  from collections import defaultdict
  def ii():
      return int(input())
  def li():
      return list(map(int,input().split()))
  if(os.path.exists('input.txt')):
      sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w") 
  else:
      input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  
  def solve():
      l,r = li()
      x = r//2 +1
      if l<=x:
          print(r%x)
      else:
          print(r%l)
  
  
  t = 1
  t = ii()
  for _ in range(t):
      solve()
  
  
  