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>
  #pragma GCC optimize(3, "Ofast", "inline")
  using namespace std;
  template <typename T>
  istream& operator>>(istream& is, vector<T>& v) {
    for (auto& it : v) is >> it;
    return is;
  }
  template <class L, class R>
  ostream& operator<<(ostream& os, pair<L, R> P) {
    return os << "(" << P.first << "," << P.second << ")";
  }
  template <class T>
  ostream& operator<<(ostream& os, vector<T> V) {
    os << "[ ";
    for (auto v : V) os << v << " ";
    return os << "]";
  }
  template <class T>
  ostream& operator<<(ostream& os, set<T> second) {
    os << "{ ";
    for (auto s : second) os << s << " ";
    return os << "}";
  }
  template <typename Arg1>
  void __f(const char* name, Arg1&& arg1) {
    cerr << name << " : " << arg1 << '\n';
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << " : " << arg1 << " |";
    __f(comma + 1, args...);
  }
  void display(vector<long long int> v) {
    for (long long int i = 0; i < v.size(); i++) cout << v[i] << " ";
    cout << '\n';
  }
  void solve() {
    long long int n, s;
    cin >> n >> s;
    long long int div = n / 2;
    div++;
    cout << s / div << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long int tt = 1;
    cin >> tt;
    while (tt--) {
      solve();
    }
  }
  