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
  
  # sys.setrecursionlimit(200005)
  int1 = lambda x: int(x)-1
  p2D = lambda x: print(*x, sep="\n")
  def II(): return int(sys.stdin.readline())
  def LI(): return list(map(int, sys.stdin.readline().split()))
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  def LI1(): return list(map(int1, sys.stdin.readline().split()))
  def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
  def SI(): return sys.stdin.readline().rstrip()
  inf = 10**16
  md = 10**9+7
  # md = 998244353
  
  def solve():
      l,r=LI()
      l=max(l,r//2+1)
      print(r%l)
  
  for testcase in range(II()):
      solve()
  