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 Type>
  void RIP(Type x) {
    cout << x << "\n";
  }
  template <typename Type, typename... Targs>
  void RIP(Type x, Targs... args) {
    cout << x << " ";
    RIP(args...);
  }
  mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
  const int N = 1e6 + 10;
  const double eps = 1e-9;
  const int inf = 0x3f3f3f3f;
  const long long INF = 0x3f3f3f3f3f3f3f3f;
  const int mo = (1 ? 998244353 : 1000000007);
  long long mul(long long a, long long b, long long m = mo) { return a * b % m; }
  long long fpow(long long a, long long b, long long m = mo) {
    long long ans = 1;
    for (; b; a = mul(a, a, m), b >>= 1)
      if (b & 1) ans = mul(ans, a, m);
    return ans;
  }
  inline long long read() {
    long long x = 0, tag = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar())
      if (c == '-') tag = -1;
    for (; isdigit(c); c = getchar()) x = x * 10 + c - 48;
    return x * tag;
  }
  const double pi = acos(-1.0);
  double readf() {
    double x;
    if (scanf("%lf", &x) != 1) exit(0);
    return x;
  }
  template <typename T>
  T sqr(T x) {
    return x * x;
  }
  long long a[N];
  void Solve() {
    long long n = read(), m = read();
    cout << m / (n + 1 - (n + 1 >> 1)) << "\n";
  }
  int main() {
    int T = 1;
    T = read();
    for (int kase = (1); kase <= (T); ++kase) {
      Solve();
    }
    return 0;
  }
  