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.io.*;
  public class Main {
  	public static void main(String[] args) throws IOException, InterruptedException {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		int jj = sc.nextInt();
  		while(jj-->0) {
  			long n = sc.nextLong();
  			long s = sc.nextLong();
  			
  				long div  = (n/2)+1;
  				if(n == 1) {
  					pw.println(s);
  				}
  				else {
  					if(s%div==0) {
  						
  						pw.println(s/div);
  					}
  					else {
  						//pw.println("M");
  						pw.println(s/div);
  					}
  				}
  				
  				
  			
  		}
  				
  		
  		pw.flush();
  	}
  	
  	
  }
  
  class edge implements Comparable<edge> {
  	int from;
  	int dest;
  	double cost;
  
  	public edge(int from, int dest, double cost) {
  		this.from = from;
  		this.dest = dest;
  		this.cost = cost;
  
  	}
  public String toString() {
  	return from+" "+dest+" "+cost;
  }
  	@Override
  	public int compareTo(edge o) {
  //		return cost - o.cost;
  		if (cost < o.cost)
  			return -1;
  		if (cost > o.cost)
  			return 1;
  		return 0;
  	}
  
  }
  
  class pair {
  	int a;
  	int b;
  
  	public pair(int a, int b) {
  		this.a = a;
  		this.b = b;
  	}
  
  }
  
  class Scanner {
  	StringTokenizer st;
  	BufferedReader br;
  
  	public Scanner(InputStream s) {
  		br = new BufferedReader(new InputStreamReader(s));
  	}
  
  	public String next() throws IOException {
  		while (st == null || !st.hasMoreTokens())
  			st = new StringTokenizer(br.readLine());
  		return st.nextToken();
  	}
  
  	public int nextInt() throws IOException {
  		return Integer.parseInt(next());
  	}
  
  	public long nextLong() throws IOException {
  		return Long.parseLong(next());
  	}
  
  	public String nextLine() throws IOException {
  		return br.readLine();
  	}
  
  	public double nextDouble() throws IOException {
  		String x = next();
  		StringBuilder sb = new StringBuilder("0");
  		double res = 0, f = 1;
  		boolean dec = false, neg = false;
  		int start = 0;
  		if (x.charAt(0) == '-') {
  			neg = true;
  			start++;
  		}
  		for (int i = start; i < x.length(); i++)
  			if (x.charAt(i) == '.') {
  				res = Long.parseLong(sb.toString());
  				sb = new StringBuilder("0");
  				dec = true;
  			} else {
  				sb.append(x.charAt(i));
  				if (dec)
  					f *= 10;
  			}
  		res += Long.parseLong(sb.toString()) / f;
  		return res * (neg ? -1 : 1);
  	}
  
  	public boolean ready() throws IOException {
  		return br.ready();
  	}
  }