description: You are given two positive integers n and s. Find the maximum possible median of an array of n non-negative integers (not necessarily distinct), such that the sum of its elements is equal to s.
  
  A median of an array of integers of length m is the number standing on the ⌈ {m/2} ⌉-th (rounding up) position in the non-decreasing ordering of its elements. Positions are numbered starting from 1. For example, a median of the array [20,40,20,50,50,30] is the ⌈ m/2 ⌉-th element of [20,20,30,40,50,50], so it is 30. There exist other definitions of the median, but in this problem we use the described definition.
  
  Input
  
  The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
  
  Each test case contains a single line with two integers n and s (1 ≤ n, s ≤ 10^9) — the length of the array and the required sum of the elements.
  
  Output
  
  For each test case print a single integer — the maximum possible median.
  
  Example
  
  Input
  
  
  8
  1 5
  2 5
  3 5
  2 1
  7 17
  4 14
  1 1000000000
  1000000000 1
  
  
  Output
  
  
  5
  2
  2
  0
  4
  4
  1000000000
  0
  
  Note
  
  Possible arrays for the first three test cases (in each array the median is underlined):
  
    * In the first test case [\underline{5}] 
    * In the second test case [\underline{2}, 3] 
    * In the third test case [1, \underline{2}, 2] 
solution: import queue
  from math import *
  import sys
  from collections import *
  from random import *
  from itertools import *
  
  sys.setrecursionlimit(99999)
  eps = sys.float_info.epsilon
  P = 2
  INF = 1e9 + 1
  MOD = 1000000007
  
  
  def is_prime(n):
      if n == 0 or n == 1:
          return False
      d = 2
      while d * d <= n:
          if n % d == 0:
              return False
          d += 1
      return True
  
  
  def div_up(n, k):
      if n % k == 0:
          return n // k
      else:
          return n // k + 1
  
  
  def num_len(n, base = 10):
      if n == 0:
          return 1
      return int(floor(log(n, base) + 1))
  
  
  def dfs_sample(graph, cl, p, v):
      cl[v] = 1
      for to in graph[v]:
          if cl[to] == 1 and p[v] != to:
              # yay, cycle
              pass
          elif cl[to] == 0:
              p[to] = v
              dfs_sample(graph, cl, p, to)
      cl[v] = 2
  
  
  def down(a):
      for i in range(1, len(a)):
          if a[i] > a[i - 1]:
              return False
      return True
  
  
  def up(a):
      for i in range(1, len(a)):
          if a[i] < a[i - 1]:
              return False
      return True
  
  
  def code(c):
      return ord(c) - 32
  
  
  def _hash_(s):
      res, p = 0, 1
      for i in range(len(s)):
          res += (code(s[i]) * p)
          res %= MOD
          p *= P
          p %= MOD
      return res % MOD
  
  
  def remove_edge(v, u, graph):
      graph[v].remove(u)
      graph[u].remove(v)
  
  
  b = True
  def dfs_stack(start, graph, visited, flags):
      global b
      stack = []
      visited[start] = True
      stack.append(start)
      while len(stack) != 0:
          v = stack.pop()
          if not flags[v]:
              b = False
          for to in graph[v]:
              if not visited[to]:
                  visited[to] = True
                  stack.append(to)
  
  
  def all_eq(a):
      for i in range(len(a) - 1):
          if a[i] != a[i + 1]:
              return False
      return True
  
  
  def solve():
      n, s = map(int, input().split())
      if n == 1:
          print(s)
      else:
          if n % 2 != 0:
              print(2 * s // (n + 1))
          else:
              print(2 * s // (n + 2))
  
  
  for _ in range(int(input())):
      solve()
  
  
  def debug():
      pass
  
  
  # debug()