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: import java.util.*;
  import java.io.*;
  public class Main {
  // For fast input output
  static class FastReader {
  BufferedReader br;
  StringTokenizer st;
  public FastReader()
  { try {br = new BufferedReader(
  new FileReader("input.txt"));
  PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
  System.setOut(out);}
  catch(Exception e) { br = new BufferedReader(new InputStreamReader(System.in));}
  }
  String next()
  {
  while (st == null || !st.hasMoreElements()) {
  try {st = new StringTokenizer(br.readLine());}
  catch (IOException e) {
  e.printStackTrace();}
  }
  return st.nextToken();
  }
  int nextInt() { return Integer.parseInt(next()); }
  long nextLong() { return Long.parseLong(next()); }
  double nextDouble() {return Double.parseDouble(next()); }
  String nextLine()
  {
  String str = "";
  try {
  str = br.readLine();
  }
  catch (IOException e) {
  e.printStackTrace();
  }
  return str;
  }
  }
  // end of fast i/o code
  public static void main(String[] args) {
    FastReader reader = new FastReader();
    int Q = reader.nextInt();
    outer: for (int q = 0; q < Q; q++) {
      int n = reader.nextInt(), sum = reader.nextInt();
      int index = (n % 2 == 1 ? n / 2 : (n - 1) / 2);
      // sum -= index;
      System.out.println(sum / (n - index));
    }
  }}