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.util.*;
  import java.lang.*;
  import java.io.*;
  
  public class A_Median_Maximization
  {
  	public static void main (String[] args) throws java.lang.Exception
  	{
  		// your code goes here
  		
  		FastReader s = new FastReader();
  		
  		int t = s.nextInt();
  		
  		while(t-- != 0){
  		    long n = s.nextLong();
              long sum = s.nextLong();
  
              long l = -1;
              long r = 1000000001;
  
              while(r-l > 1){
                  // System.out.println(r+" "+l);
                  if(ans(l,r,n,sum) == 0){
                      r = (r+l)/2;
                  }
                  else{
                      l = (r+l)/2;
                  }
              }
  
              System.out.println(l);
  			
  		}
  	}
  
      public static int ans(long l,long r, long n, long sum){
          long m = (r+l)/2;
  
          // System.out.println(((n/2)+1)*m);
  
          long sol = ((n/2)+1)*m;
  
          // System.out.println(sol);
  
          if(sol <= sum){
              return 1;
          }
          else{
              return 0;
          }
      }
  	
  	public static int ans(long l, long r, long v){
  		
  //		System.out.println((r-l)/2);
  		
  		long m = (r+l)/2;
  		
  //		System.out.println(m);
  		
  		if(((46*(m*m))+(6*m)+4)>=v){
  			return 1;
  		}
  		else{
  			return 0;
  		}
  	}
  	
  	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; 
          } 
      } 
  }
  
  