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
  input = sys.stdin.readline # for strings
  # input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline # for non-strings
  
  PI = 3.141592653589793238460
  INF =  float('inf')
  MOD  = 1000000007
  # MOD = 998244353
  
  
  def bin32(num):
      return '{0:032b}'.format(num)
  
  def add(x,y):
      return (x+y)%MOD
  
  def sub(x,y):
      return (x-y+MOD)%MOD
  
  def mul(x,y):
      return (x*y)%MOD
  
  def gcd(x,y):
      if y == 0:
          return x
      return gcd(y,x%y)
  
  def lcm(x,y):
      return (x*y)//gcd(x,y)
  
  def power(x,y):
      res = 1
      x%=MOD
      while y!=0:
          if y&1 :
              res = mul(res,x)
          y>>=1
          x = mul(x,x)
          
      return res
          
  def mod_inv(n):
      return power(n,MOD-2)
  
  def prob(p,q):
      return mul(p,power(q,MOD-2))    
    
  def ii():
      return int(input())
  
  def li():
      return [int(i) for i in input().split()]
  
  def ls():
      return [i for i in input().split()]
  
  for t in range(ii()):
      t+=1
      l,r = li()
      if r%2 == 1:
          if l > r//2:
              print(r%l)
          else:
              print(r%((r+1)//2))
      else:
          x = r//2
          x+=1
          if x>=l:
              print(r%x)
          else:
              print(r%l)