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.util.*;
  
  	public class practice{
  		public static void main(String[] args){
  			FastScanner r = new FastScanner();					
  			int t = r.i() ,n,s;
  			StringBuffer ans = new StringBuffer("");
  			while(t-->0) {
  				n = r.i();
  				s = r.i();
  
  				if(n%2==0)
  				n -= (n/2 - 1);
  				else
  					n -= (n/2 );
  				
  //				p(s/n);
  				ans.append(s/n + "\n");
  				
  			}				
  			p(ans);
  		}
  
  		
  		static <T> void p(T s) {
  			System.out.println(s);
  		}
  		
  		static void p() {
  			System.out.println();
  		}		
  		static <T> void pn(T s) {
  			System.out.print(s);
  		}
  
  	}
  		
  
  	
  	class FastScanner {
  	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  	    StringTokenizer st=new StringTokenizer("");
  	    String n() {
  	        while (!st.hasMoreTokens())
  	            try {
  	                st=new StringTokenizer(br.readLine());
  	            } catch (IOException e) {
  	                e.printStackTrace();
  	            }
  	        return st.nextToken();
  	    }
  	    int[] ia(int n) {
  	    	int[] arr = new int[n];
  	    	for(int i=0;i<n;i++)
  	    		arr[i] = i();
  	    	return arr;
  	    }
  	    String[] sa(int n) {
  	    	String[] arr = new String[n];
  	    	for(int i=0;i<n;i++)
  	    		arr[i] = n();
  	    	return arr;
  	    }
  	    long[] la(int n) {
  	    	long[] arr = new long[n];
  	    	for(int i=0;i<n;i++)
  	    		arr[i] = l();
  	    	return arr;
  	    }
  	    String nl()
  	    {
  	        String str = "";
  	        try
  	        {
  	            str = br.readLine();
  	        }
  	        catch (IOException e)
  	        {
  	            e.printStackTrace();
  	        }
  	        return str;
  	    }
  
  	    int i() {
  	        return Integer.parseInt(n());
  	    }
  	    long l() {
  	        return Long.parseLong(n());
  	    }
  	}
  