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;
  struct hash_pair {
    template <class T1, class T2>
    size_t operator()(const pair<T1, T2> &p) const {
      auto hash1 = hash<T1>{}(p.first);
      auto hash2 = hash<T2>{}(p.second);
      return hash1 ^ hash2;
    }
  };
  long long int n, m;
  vector<long long int> adj[1];
  bool visited[1];
  long long int start[1];
  long long int end[1];
  long long int parent[1];
  long long int level[1];
  vector<long long int> dfs_order;
  void dfs(long long int node) {
    visited[node] = true;
    for (long long int next : adj[node]) {
      if (!visited[next]) {
        parent[next] = node;
        level[next] = level[node] + 1;
        dfs(next);
      }
    }
  }
  long long int dist[1];
  void bfs(long long int start) {
    memset(dist, -1, sizeof dist);
    queue<long long int> q;
    dist[start] = 0;
    q.push(start);
    while (!q.empty()) {
      long long int v = q.front();
      q.pop();
      for (long long int e : adj[v]) {
        if (dist[e] == -1) {
          dist[e] = dist[v] + 1;
          q.push(e);
        }
      }
    }
  }
  long long int lift(long long int a, long long int dist,
                     vector<vector<long long int>> &up) {
    for (long long int i = 0; i < 20; i++) {
      if (dist & (1 << i)) {
        a = up[a][i];
      }
    }
    return a;
  }
  void preprocesslift(vector<vector<long long int>> &up) {
    for (long long int j = 1; j < 20; j++) {
      for (long long int i = 1; i <= n; i++) {
        if (up[i][j - 1] != -1) {
          up[i][j] = up[up[i][j - 1]][j - 1];
        }
      }
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t;
    cin >> t;
    while (t > 0) {
      long long int n, s;
      cin >> n >> s;
      if (((n / 2) + 1) > s)
        cout << "0\n";
      else if (s == 1)
        cout << n << "\n";
      else {
        long long int ans = 0;
        if (n % 2 == 0)
          ans = s / ((n / 2) + 1);
        else
          ans = s / ((n / 2) + 1);
        cout << ans << "\n";
      }
      t--;
    }
    return 0;
  }
  