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;
  bool isPrime(long long n) {
    if (n == 1) return false;
    if (n == 2) return true;
    for (long long i = 2; i * i <= n; i++) {
      if (n % i == 0) return false;
    }
    return true;
  }
  const long long N = 1e7 + 5;
  bool sieve[N];
  void soe() {
    for (long long i = 2; i < N; ++i) sieve[i] = true;
    for (long long i = 2; i * i < N; ++i) {
      if (sieve[i] == true) {
        for (long long j = i * i; j < N; j += i) {
          sieve[j] = false;
        }
      }
    }
  }
  long long countDivisors(long long n) {
    long long cnt = 0;
    for (long long i = 1; i <= sqrt(n); i++) {
      if (n % i == 0) {
        if (n / i == i)
          cnt++;
        else
          cnt = cnt + 2;
      }
    }
    return cnt;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long fact(long long n);
  long long nCr(long long n, long long r) {
    return fact(n) / (fact(r) * fact(n - r));
  }
  long long fact(long long n) {
    long long res = 1;
    for (long long i = 2; i <= n; i++) res = res * i;
    return res;
  }
  long long PowMod(long long n) {
    long long ret = 1;
    long long a = 2;
    while (n > 0) {
      if (n & 1) ret = ret * a % 1000000007;
      a = a * a % 1000000007;
      n >>= 1;
    }
    return ret;
  }
  signed main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL);
    ;
    long long t = 1;
    cin >> t;
    while (t--) {
      long long n, s;
      cin >> n >> s;
      long long div = 1 + n / 2;
      cout << s / div << endl;
    }
    return 0;
  }
  