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: /*
  
  Sample Input 1
  
  8
  1 5
  2 5
  3 5
  2 1
  7 17
  4 14
  1 1000000000
  1000000000 1
  
  Sample Output 1
  
  5
  2
  2
  0
  4
  4
  1000000000
  0
  
  */
  import java.io.*;
  import java.util.*;
    
  public class R1566A
  {
      static int MAX(int x, int y){return ((x > y) ? x : y );}
      static int MIN(int x, int y){return ((x < y) ? x : y );}
      static int sum(int[] A){int sum = 0;for(int i=0; i<A.length; i++)sum+=A[i];return (sum);}
      static int gcd(int a, int b){if (b == 0)return a;return gcd(b, a % b);}
      static boolean isPowerOfTwo(int n){if(n==0)return false;return (int)(Math.ceil((Math.log(n) / Math.log(2)))) ==(int)(Math.floor(((Math.log(n) / Math.log(2)))));}
      static int countDigits(long l)
      {
              if (l >= 1000000000000000000L) return 19;
              if (l >= 100000000000000000L) return 18;
              if (l >= 10000000000000000L) return 17;
              if (l >= 1000000000000000L) return 16;
              if (l >= 100000000000000L) return 15;
              if (l >= 10000000000000L) return 14;
              if (l >= 1000000000000L) return 13;
              if (l >= 100000000000L) return 12;
              if (l >= 10000000000L) return 11;
              if (l >= 1000000000L) return 10;
              if (l >= 100000000L) return 9;
              if (l >= 10000000L) return 8;
              if (l >= 1000000L) return 7;
              if (l >= 100000L) return 6;
              if (l >= 10000L) return 5;
              if (l >= 1000L) return 4;
              if (l >= 100L) return 3;
              if (l >= 10L) return 2;
              return 1;
      }
      static void sort2D(int[][] Arr){Arrays.sort(Arr, (a, b) -> Double.compare(a[0], b[0]));}
      public static void main(String[] args) throws IOException
      {
          BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
          if (System.getProperty("ONLINE_JUDGE") == null)
          {
              try 
              {
                  in = new BufferedReader(new FileReader("input.txt"));
                  System.setOut(new PrintStream(new FileOutputStream("output.txt")));
              }
              catch (Exception e) {
                  //System.out.println("Exception");
              }
          }
          int T = Integer.parseInt(in.readLine());
          while( T -- > 0)
          {
              String lines = in.readLine();
              String[] strs = lines.trim().split("\\s+");
              int n = Integer.parseInt(strs[0]);
              int s = Integer.parseInt(strs[1]);
              int temp = 1;
              if(n == 1)
                  System.out.println(s);
              else if(s == 1)
                  System.out.println("0");
              else
              {
                  if(n % 2 == 0)
                      temp = n / 2 + 1;
                  else
                      temp = (int)Math.ceil(n/(2*1.0));
                  temp = s/temp;
                  System.out.println(temp);
              }
          }
      }
  }