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.IOException;
  import java.io.InputStreamReader;
  import java.sql.Array;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.StringTokenizer;
  
  public class codeforcesb {
      public static class Pair {
          long f;
          long c;
  
          Pair(long f, long c) {
              this.f = f;
              this.c = c;
          }
      }
  
      public static void main(String[] args) {
  
  
              FastScanner fs = new FastScanner();
              int T = fs.nextInt();
  
  
              for (int tt = 0; tt < T; tt++) {
                 int n=fs.nextInt(),s=fs.nextInt();
                 int m=(n/2)+1;
                  System.out.println(s/m);
              }
  
  
  }
  
  
  
  
  
      static void sort(int[] a) {
          ArrayList<Integer> l=new ArrayList<>();
          for (int i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static long power(long a, long b) {
          if(b == 0) return 1l;
          long temp = power(a, b / 2);
          temp = (temp * temp) ;
          return b % 2 == 0 ? temp : (temp * a);
      }
  
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long[] readLongArray(int n) {
              long [] a = new long[n];
              a[0] = 0l;
              for (int i = 0; i <n; i++) a[i] = nextLong();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  
  
  }
  
  