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;
  const long long MOD = 1000000007;
  const int MAX_INT = 2147483647;
  const long long MAX_LL = (long long)(9e18);
  const int N = 100005;
  int n, m, q, ui, vi, tmp = 0, cnt = 0, l, r, mid, tmp2;
  char s[N];
  int k, ans = 0, sum = 0;
  bool flag;
  void solve(int CASE) {
    scanf("\n%d%d", &n, &k);
    if (n == 1)
      printf("%d\n", k);
    else if (n == 2) {
      printf("%d\n", k / n);
    } else
      printf("%d\n", k / (n - ((n + 1) / 2 - 1)));
    return;
  }
  int main() {
    int __t = 1;
    scanf("\n%d", &__t);
    for (int _T = 1; _T <= __t; _T++) {
      solve(_T);
    }
    return 0;
  }
  