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.IOException;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.Scanner;
  import java.util.StringTokenizer;
  
  public class Main {
  
      public static void main(String[] args){
  
          FastReader fastReader = new FastReader();
  
          long t = fastReader.nextLong();
  
          while (t-- >0){
  
              long n = fastReader.nextLong();
              long s = fastReader.nextLong();
  
  //            float rf = (float) n/(float) 2;
  
              long roundoff = 0 ;
              double ans = 0;
  
              if(n%2==0){
  
                  roundoff = n/2;
  
              }
              else {
  
                  roundoff = (n/2)+ 1 ;
  
              }
  
  //            long ans = s / roundoff ;
              ans = Math.floor( (double) s / (double) (n-roundoff+1) );
  
              System.out.println((long) ans);
  
  
          }
  
  
  
  
  
  
  
  
      }
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
              static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader()
          {
              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;
          }
      }
  
  
  
      }
  
  