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: from collections import deque, Counter
  from collections import defaultdict as dfd
  from bisect import bisect, bisect_left
  from math import sqrt, gcd, ceil, factorial
  from heapq import heapify, heappush, heappop
  
  import sys
  sys.setrecursionlimit(10**7) 
  
  MOD = 10**9 + 7
  inf = float("inf")
  ans_ = []
  
  def nin():return int(input())
  def ninf():return int(file.readline())
  
  def st():return (input().strip())
  def stf():return (file.readline().strip())
  
  def read(): return list(map(int, input().strip().split()))
  def readf():return list(map(int, file.readline().strip().split()))
  
  def readfl(): return list(map(float, input().strip().split()))
  def readffl():return list(map(float, file.readline().strip().split()))
  
  
  
  
  
  # file = open("input.txt", "r")
  def solve():
      for _ in range(nin()):
          l, r = read()
          x = max(l, r//2 + 1)
          ans_.append(r%x)
      
  solve()
  
  for i in ans_:
      print(i)
      