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;
  long long gcd(long long a, long long b) {
    if (a < 0) a = -a;
    if (b < 0) b = -b;
    return b == 0 ? a : gcd(b, a % b);
  }
  template <typename T>
  void read(T &res) {
    bool flag = false;
    char ch;
    while (!isdigit(ch = getchar())) (ch == '-') && (flag = true);
    for (res = ch - 48; isdigit(ch = getchar());
         res = (res << 1) + (res << 3) + ch - 48)
      ;
    flag && (res = -res);
  }
  long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
  long long qp(long long a, long long b, long long mod) {
    long long ans = 1;
    if (b == 0) {
      return ans % mod;
    }
    while (b) {
      if (b % 2 == 1) {
        b--;
        ans = ans * a % mod;
      }
      a = a * a % mod;
      b = b / 2;
    }
    return ans % mod;
  }
  long long qpn(long long a, long long b, long long p) {
    long long ans = 1;
    a %= p;
    while (b) {
      if (b & 1) {
        ans = (ans * a) % p;
        --b;
      }
      a = (a * a) % p;
      b >>= 1;
    }
    return ans % p;
  }
  signed main() {
    long long t;
    read(t);
    while (t--) {
      long long n, s;
      read(n);
      read(s);
      long long ans = s / n;
      ans = s / (n / 2 + 1);
      printf("%lld\n", ans);
    }
  }
  