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: #include <bits/stdc++.h>
  template <typename Arg1>
  void __f(const char* name, Arg1&& arg1) {
    std::cerr << name << " : " << arg1 << "\n";
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');
    std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
  }
  std::pair<int, int> operator+(std::pair<int, int> a, std::pair<int, int> b) {
    return {a.first + b.first, a.second + b.second};
  }
  std::pair<long long, long long> operator+(std::pair<long long, long long> a,
                                            std::pair<long long, long long> b) {
    return {a.first + b.first, a.second + b.second};
  }
  std::ostream& operator<<(std::ostream& out, std::pair<int, int> a) {
    out << a.first << " " << a.second << "\n";
    return out;
  }
  std::ostream& operator<<(std::ostream& out, std::pair<long long, long long> a) {
    out << a.first << " " << a.second << "\n";
    return out;
  }
  std::istream& operator>>(std::istream& in, std::pair<int, int>& a) {
    in >> a.first >> a.second;
    return in;
  }
  std::istream& operator>>(std::istream& in, std::pair<long long, long long>& a) {
    in >> a.first >> a.second;
    return in;
  }
  long long bpow(long long x, long long v) {
    if (v == 0) return 1 % 1000000007;
    if (v % 2 == 1) return (x * bpow(x, v - 1)) % 1000000007;
    long long r = bpow(x, v / 2);
    return (r * r) % 1000000007;
  }
  using namespace std;
  void solve() {
    long long n, s;
    cin >> n >> s;
    long long a = n - n / 2 + (1 - n % 2);
    cout << s / a << "\n";
  }
  int main() {
    std::ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
  }
  