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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  long long M = 1e9 + 7;
  long long mod(long long x) { return ((x % M + M) % M); }
  long long gcd(long long a, long long b);
  long long lcm(long long a, long long b);
  bool prime(long long n);
  long long BinExp(long long base, long long power);
  long long ModInverse(long long base);
  long long nCr(long long n, long long r);
  long long fact(long long n) { return (n <= 1) ? 1 : n * fact(n - 1); }
  void solve() {
    long long n, s;
    cin >> n >> s;
    n -= (n - 1) / 2;
    cout << (long long)(s / n) << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) solve();
  }
  long long nCr(long long n, long long r) {
    r = (n - r <= r) ? n - r : r;
    long long ans = 1;
    for (long long i = 0; i < r; i++) {
      ans *= (n - i);
      ans /= (i + 1);
    }
    return ans;
  }
  long long gcd(long long a, long long b) {
    if (a == 0) return b;
    return (a == 1) ? a : gcd(b % a, a);
  }
  long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
  bool prime(long long n) {
    if (n < 2) return false;
    for (long long i = 2; i <= sqrt(n); i++)
      if (n % i == 0) return false;
    return true;
  }
  long long BinExp(long long base, long long power) {
    if (!power) return 1;
    long long res = 1;
    while (power > 1) {
      if (power % 2) {
        power--;
        res *= base;
      } else {
        base *= base;
        power /= 2;
      }
    }
    return base * res;
  }
  long long ModInverse(long long base) {
    long long power = M - 2;
    if (!power) return 1;
    long long res = 1;
    while (power > 1) {
      if (power % 2) {
        power--;
        res = mod(res * base);
      } else {
        base = mod(base * base);
        power /= 2;
      }
    }
    return mod(base * res);
  }
  