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 Counter, defaultdict, deque
  import bisect
  from sys import stdin, stdout
  from itertools import repeat
  import math
  
  
  def inp(force_list=False):
      re = map(int, raw_input().split())
      if len(re) == 1 and not force_list:
          return re[0]
      return re
  
  def inst():
      return raw_input().strip()
  
  def gcd(x, y):
     while(y):
         x, y = y, x % y
     return x
  
  mod = 1000000007
  
  def my_main():
      kase = inp()
      pans = []
      for _ in range(kase):
          l, r = inp()
          if r>=l*2:
              ans = r%(r/2+1)
          else:
              ans = r%l
  
  
  
          pans.append(str(ans))
  
          # pans.append(''.join(map(str, reversed(ans))))
      print '\n'.join(pans)
  
  
      # print '\n'.join(pans)
  
  my_main()
  