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 os
  import sys
  from io import BytesIO, IOBase
  #import sys
  #input = sys.stdin.buffer.readline
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  #######################################
  from itertools import permutations
  from collections import *
  from collections import deque
  from operator import itemgetter , attrgetter
  from decimal import *
  import bisect
  import math
  import heapq as hq
  #import sympy
  MOD=10**9 +7
  
  
  def is_prime(n):
    if n == 2 or n == 3: return True
    if n < 2 or n%2 == 0: return False
    if n < 9: return True
    if n%3 == 0: return False
    r = int(n**0.5)
    # since all primes > 3 are of the form 6n ± 1
    # start with f=5 (which is prime)
    # and test f, f+2 for being prime
    # then loop by 6.
    f = 5
    while f <= r:
  
      if n % f == 0: return False
      if n % (f+2) == 0: return False
      f += 6
    return True
  
  def pow(a,b,m):
      ans=1
      while b:
          if b&1:
              ans=(ans*a)%m
          b//=2
          a=(a*a)%m
      return ans
  #vis=[]
  #graph=[]
  
  def ispalindrome(s):
      if s[:]==s[::-1]:
          return 1
      return 0
  dp=[]
  limit=[]
  v=[]
  def dpdfs(u,t=-1):
      dp[0][u]=0
      dp[1][u]=0
      for i in v[u]:
          if i==t:
  
              continue
          if dp[1][i]==-1:
              dpdfs(i,u)
          dp[0][u]+=max(abs(limit[0][u]-limit[1][i])+dp[1][i],abs(limit[0][u]-limit[0][i])+dp[0][i])
          dp[1][u] += max(abs(limit[1][u] - limit[1][i]) + dp[1][i], abs(limit[1][u] - limit[0][i]) + dp[0][i])
  vis=[]
  f=0
  def dfs(i):
      vis[i]=1
      act[i]=1
      for j in v[i]:
          if act[j]:
              f=1
              #print(-1)
              return -1
          if vis[j]==0:
             if dfs(j)==-1:
                 return -1
      act[i]=0
      ans.append(i)
      return 0
  from queue import PriorityQueue
  
  def z_algorithm(s):
      res = [0] * len(s)
      res[0] = len(s)
      i, j = 1, 0
      while i < len(s):
          while i + j < len(s) and s[j] == s[i + j]:
              j += 1
          res[i] = j
          if j == 0:
              i += 1
              continue
          k = 1
          while i + k < len(s) and k + res[k] < j:
              res[i + k] = res[k]
              k += 1
          i, j = i + k, j - k
      return res
  def gcd(a, b):
      if a == 0:
          return b
      return gcd(b % a, a)
  
  
  # Function to return LCM of two numbers
  def lcm(a, b):
      return (a / gcd(a, b)) * b
  
  
  def gcdExtended(a, b):
      # Base Case
      if a == 0:
          return b, 0, 1
  
      gcd, x1, y1 = gcdExtended(b % a, a)
  
      # Update x and y using results of recursive
      # call
      x = y1 - (b // a) * x1
      y = x1
  
      return gcd, x, y
  import re
  for _ in range(int(input())):
      l,r=map(int,input().split())
      t=1+(r//2)
      if t>=l:
          print(r%t)
      else:
          print(r%l)
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  