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.io.BufferedReader;
  import java.io.BufferedWriter;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.OutputStreamWriter;
  import java.util.ArrayList;
  import java.util.Arrays;
  
  
  
  
  public class Codeforces {
   public static void main(String[] args) throws IOException{
  
         BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
         BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
         
         long t = Long.parseLong(br.readLine().trim());
          
         while(t-->0){
               String arr[] = br.readLine().split(" ");
               long s = Long.parseLong(arr[1].trim());
               long n = Long.parseLong(arr[0].trim());
                long ans =0;
               if(n==1){System.out.println(s);
              }
              else if(s==1){
                  System.out.println("0");
              }
              else{
                if(n%2==0){
                  ans = (s/((n/2)+1));
                }
                else{
                    ans = (s/(n/2+1));
                }
                System.out.println(ans);
              }
              }
         br.close();
         log.flush();
     } 
  
  static void print(int arr[]) {
      System.out.print("Print :- ");
      for(int i=0 ; i<arr.length ; i++) {
          System.out.print(arr[i]+" ");
      }
      System.out.println();
          }   // end-print
  
  
  static boolean isPrime(long n){
      if (n <= 1)  return false;
  
      else if (n == 2) return true;
  
      else if (n % 2 == 0)return false;
  
      for (int i = 3; i <= Math.sqrt(n); i += 2){
          if (n % i == 0) return false;
      }
      return true;
          }//end isprime
   
  
  static boolean isPrime(char a){
      if (a == '2' || a == '3' || a == '5' || a == '7') {
          return true;
      }
  
      return false;
      
  }
  
  
  static int computeXOR(int n){
          if (n % 4 == 0)
              return n;
  
          if (n % 4 == 1)
              return 1;
        
         
          if (n % 4 == 2)
              return n + 1;
        
          return 0;
          }
          // end compute Xor
  }
  