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>
  using namespace std;
  template <typename T>
  void error(T it) {}
  template <typename Ty, typename T, typename... Args>
  void error(Ty it, T a, Args... arg) {
    cout << *it << " = " << a << '\t';
    error((++it), arg...);
  }
  void _split(string s, vector<string> &_os) {
    string se;
    long long x = 0;
    for (auto i : s) {
      if (i == ',' && x == 0)
        _os.push_back(se), se = "";
      else
        se += i;
      if (i == '(') x++;
      if (i == ')') x--;
    }
    _os.push_back(se);
  }
  long long mod = 1e+9 + 7;
  long long inf = (2e+18);
  long double pi = 3.1415926536;
  const long long N = 2e+5 + 3;
  const long long _N = 2e+6 + 3;
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long test = 1;
    cin >> test;
    while (test--) {
      long long n, s;
      cin >> n >> s;
      long long m;
      if (n % 2 == 0) {
        m = n / 2 + 1;
      } else {
        m = (n + 1) / 2;
      }
      long long t1 = s % m;
      long long t2 = (s - t1) / m;
      cout << t2 << endl;
    }
    cerr << "\nTime : "
         << 1000 * (long double)clock() / (long double)CLOCKS_PER_SEC << "ms\t";
  }
  