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 add(long long x, long long y) {
    long long res = x + y;
    return (res >= 1000000007 ? res - 1000000007 : res);
  }
  long long mul(long long x, long long y) {
    long long res = x * y;
    return (res >= 1000000007 ? res % 1000000007 : res);
  }
  long long sub(long long x, long long y) {
    long long res = x - y;
    return (res < 0 ? res + 1000000007 : res);
  }
  long long power(long long x, long long y) {
    long long res = 1;
    x %= 1000000007;
    while (y) {
      if (y & 1) res = mul(res, x);
      y >>= 1;
      x = mul(x, x);
    }
    return res;
  }
  long long mod_inv(long long x) { return power(x, 1000000007 - 2); }
  long long floorSqrt(long long x) {
    if (x == 0 || x == 1) return x;
    unsigned long long start = 1, end = x, ans;
    while (start <= end) {
      unsigned long long mid = start + (end - start) / 2;
      if (mid * mid == x) return mid;
      if (mid * mid < x) {
        start = mid + 1;
        ans = mid;
      } else
        end = mid - 1;
    }
    return ans;
  }
  string tobinary(long long n, long long length) {
    string s;
    for (long long i = 0; i < length; i++) {
      if (n & 1)
        s.push_back('1');
      else
        s.push_back('0');
      n = n >> 1;
    }
    reverse(s.begin(), s.end());
    return s;
  }
  void solve() {
    int n, s;
    cin >> n >> s;
    cout << (s) / (n / 2 + 1) << endl;
    return;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t, i = 1;
    ;
    cin >> t;
    while (t--) {
      solve();
      ++i;
    }
  }
  