description: YouKn0wWho has two even integers x and y. Help him to find an integer n such that 1 ≤ n ≤ 2 ⋅ 10^{18} and n mod x = y mod n. Here, a mod b denotes the remainder of a after division by b. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
  
  The first and only line of each test case contains two integers x and y (2 ≤ x, y ≤ 10^9, both are even).
  
  Output
  
  For each test case, print a single integer n (1 ≤ n ≤ 2 ⋅ 10^{18}) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints.
  
  Example
  
  Input
  
  
  4
  4 8
  4 2
  420 420
  69420 42068
  
  
  Output
  
  
  4
  10
  420
  9969128
  
  Note
  
  In the first test case, 4 mod 4 = 8 mod 4 = 0.
  
  In the second test case, 10 mod 4 = 2 mod 10 = 2.
  
  In the third test case, 420 mod 420 = 420 mod 420 = 0.
solution: import bisect
  import heapq
  import math
  import collections
  import sys
  import copy
  from functools import reduce
  import decimal
  from io import BytesIO, IOBase
  import os
  import itertools
  import functools
  from types import GeneratorType
  
  #
  # sys.setrecursionlimit(10 ** 9)
  decimal.getcontext().rounding = decimal.ROUND_HALF_UP
  
  graphDict = collections.defaultdict
  
  queue = collections.deque
  
  
  ################## pypy deep recursion handling ##############
  # Author = @pajenegod
  def bootstrap(f, stack=[]):
      def wrappedfunc(*args, **kwargs):
          to = f(*args, **kwargs)
          if stack:
              return to
          else:
              while True:
                  if type(to) is GeneratorType:
                      stack.append(to)
                      to = next(to)
                  else:
                      stack.pop()
                      if not stack:
                          return to
                      to = stack[-1].send(to)
  
      return wrappedfunc
  
  
  ################## Graphs ###################
  class Graphs:
      def __init__(self):
          self.graph = graphDict(set)
  
      def add_edge(self, u, v):
          self.graph[u].add(v)
          self.graph[v].add(u)
  
      def dfs_utility(self, nodes, visited_nodes, colors, parity, level):
          global count
          if nodes == 1:
              colors[nodes] = -1
          else:
              if len(self.graph[nodes]) == 1 and parity % 2 == 0:
                  if q == 1:
                      colors[nodes] = 1
                  else:
                      colors[nodes] = -1
                      count += 1
              else:
                  if parity % 2 == 0:
                      colors[nodes] = -1
                  else:
                      colors[nodes] = 1
          visited_nodes.add(nodes)
          for neighbour in self.graph[nodes]:
              new_level = level + 1
              if neighbour not in visited_nodes:
                  self.dfs_utility(neighbour, visited_nodes, colors, level - 1, new_level)
  
      def dfs(self, node):
          Visited = set()
          color = collections.defaultdict()
          self.dfs_utility(node, Visited, color, 0, 0)
          return color
  
      def bfs(self, node, f_node):
          count = float("inf")
          visited = set()
          level = 0
          if node not in visited:
              queue.append([node, level])
              visited.add(node)
          flag = 0
          while queue:
              parent = queue.popleft()
              if parent[0] == f_node:
                  flag = 1
                  count = min(count, parent[1])
              level = parent[1] + 1
              for item in self.graph[parent[0]]:
                  if item not in visited:
                      queue.append([item, level])
                      visited.add(item)
          return count if flag else -1
          return False
  
  
  ################### Tree Implementaion ##############
  class Tree:
      def __init__(self, data):
          self.data = data
          self.left = None
          self.right = None
  
  
  def inorder(node, lis):
      if node:
          inorder(node.left, lis)
          lis.append(node.data)
          inorder(node.right, lis)
      return lis
  
  
  def leaf_node_sum(root):
      if root is None:
          return 0
      if root.left is None and root.right is None:
          return root.data
      return leaf_node_sum(root.left) + leaf_node_sum(root.right)
  
  
  def hight(root):
      if root is None:
          return -1
      if root.left is None and root.right is None:
          return 0
      return max(hight(root.left), hight(root.right)) + 1
  
  
  ################## Union Find #######################
  class UF:
      """An implementation of union find data structure.
      It uses weighted quick union by rank with path compression.
      """
  
      def __init__(self, N):
          """Initialize an empty union find object with N items.
  
          Args:
              N: Number of items in the union find object.
          """
  
          self._id = list(range(N))
          self._count = N
          self._rank = [0] * N
  
      def find(self, p):
          """Find the set identifier for the item p."""
  
          id = self._id
          while p != id[p]:
              p = id[p]  # Path compression using halving.
          return p
  
      def count(self):
          """Return the number of items."""
  
          return self._count
  
      def connected(self, p, q):
          """Check if the items p and q are on the same set or not."""
  
          return self.find(p) == self.find(q)
  
      def union(self, p, q):
          """Combine sets containing p and q into a single set."""
  
          id = self._id
          rank = self._rank
  
          i = self.find(p)
          j = self.find(q)
          if i == j:
              return
  
          self._count -= 1
          if rank[i] < rank[j]:
              id[i] = j
          elif rank[i] > rank[j]:
              id[j] = i
          else:
              id[j] = i
              rank[i] += 1
  
      def add_roads(self):
          return set(self._id)
  
      def __str__(self):
          """String representation of the union find object."""
          return " ".join([str(x) for x in self._id])
  
      def __repr__(self):
          """Representation of the union find object."""
          return "UF(" + str(self) + ")"
  
  
  #################################################
  
  def rounding(n):
      return int(decimal.Decimal(f'{n}').to_integral_value())
  
  
  def factors(n):
      return set(reduce(list.__add__,
                        ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
  
  
  def p_sum(array):
      return list(itertools.accumulate(array))
  
  
  def base_change(nn, bb):
      if nn == 0:
          return [0]
      digits = []
      while nn:
          digits.append(int(nn % bb))
          nn //= bb
      return digits[::-1]
  
  
  def diophantine(a: int, b: int, c: int):
      d, x, y = extended_gcd(a, b)
      r = c // d
      return r * x, r * y
  
  
  @bootstrap
  def extended_gcd(a: int, b: int):
      if b == 0:
          d, x, y = a, 1, 0
      else:
          (d, p, q) = yield extended_gcd(b, a % b)
          x = q
          y = p - q * (a // b)
  
      yield d, x, y
  
  
  ######################################################################################
  
  '''
  Knowledge and awareness are vague, and perhaps better called illusions.
  Everyone lives within their own subjective interpretation.
                                                              ~Uchiha Itachi
  '''
  
  ################################ <fast I/O> ###########################################
  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, **kwargs):
          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)
  
  
  #############################################<I/O Region >##############################################
  
  
  def inp():
      return sys.stdin.readline().strip()
  
  
  def map_inp(v_type):
      return map(v_type, inp().split())
  
  
  def list_inp(v_type):
      return list(map_inp(v_type))
  
  
  ######################################## Solution ####################################
  
  
  for _ in range(int(inp())):
      x, y = map_inp(int)
      if x > y:
          n = 100000000 * x + y
      else:
          n = y - y % x / 2
      print(int(n))
  