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.*;
  import java.math.*;
  import java.security.*;
  import java.text.*;
  import java.time.LocalTime;
  import java.util.*;
  import java.util.concurrent.*;
  import java.util.regex.*;
  
  
  public class A {
  	
  
  	public static void main(String[] args) {
      	FastScanner scan = new FastScanner();
      	PrintWriter out = new PrintWriter(System.out);
      	
      	int T = 1;
      	T = scan.nextInt();
      		for(int tt=0; tt<T; tt++) {
      			int n = scan.nextInt(), s = scan.nextInt();
      			int num = n - ((n-1) / 2);
      			System.out.println(s / num);
      		}
  	}
  	
  	
  	
  	/*
  	static class IntComparator implements Comparator<Integer> {
  
  		public int compare(Integer o1, Integer o2) {
  		//	return Integer.compare(a[o1], a[o2]);
  			return 0;
  		}
  	}
  	*/
  	
  	
  	public static void printIntArray(int [] a) {
  		for(int i: a) System.out.print(i + " ");
  		System.out.println();
  	}
  	
  	public static void swapInt(int a, int b) {
  		int temp = a;
  		a = b;
  		b = temp;
  	}
  
  	
  	  public static void sort(int [] a) {
  	    	ArrayList<Integer> b = new ArrayList<>();
  	    	for(int i: a) b.add(i);
  	    	Collections.sort(b);
  	    	for(int i=0; i<a.length; i++) a[i]= b.get(i);
  	    }
  	  
  	  public static void reverse(int [] a) {
  	    	ArrayList<Integer> b = new ArrayList<>();
  	    	for(int i: a) b.add(i);
  	    	Collections.reverse(b);
  	    	for(int i=0; i<a.length; i++) a[i]= b.get(i);
  	    }
  
  
  		
  	
  	
      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 nextLong() {
      		return Long.parseLong(next());
      	}
      	
      	
      }
      
      
     
  }
  
  