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 {
  	static final int M = 1000000007;
  
  	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());
  		}
  
  		long[] readArrayLong(int n) {
  			long[] a = new long[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextLong();
  			return a;
  		}
  
  		int[] readArrayInt(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());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  
  	static class Pair<F, S> {
  		F first;
  		S second;
  
  		Pair(F first, S second) {
  			this.first = first;
  			this.second = second;
  		}
  
  		public F getFirst() {
  			return this.first;
  		}
  
  		public S getSecond() {
  			return this.second;
  		}
  		// Sort on basis of first term, reverse p1.first and p2.first to sort in desc
  		// Collections.sort(a, (p1, p2) -> (p1.first - p2.first));
  
  		// Sort on the basis of second term, reverse p1.second and p2.second to sort
  		// in
  		// desc
  		// Collections.sort(a, (p1, p2) -> (p1.second - p2.second));
  	}
  
  	static long __gcd(long a, long b) {
  		if (b == 0)
  			return a;
  		return __gcd(b, a % b);
  	}
  
  	/* Iterative Function to calculate (x^y) in O(log y) */
  	static long power(long a, long b) {
  		if (b == 0)
  			return 1;
  		if (b % 2 == 1)
  			return (a * power((a * a) % M, b / 2)) % M;
  		return power((a * a) % M, b / 2);
  	}
  
  	static long[] sum, a;
  
  	public static void main(String[] args) throws IOException {
  		FastScanner fs = new FastScanner();
  		int T = fs.nextInt();
  		// int T = 1;
  		StringBuffer res = new StringBuffer();
  
  		for (int tt = 1; tt <= T; tt++) {
  			int n = fs.nextInt(), s = fs.nextInt();
  			int low = n / 2 + 1;
  			res.append(s / low).append("\n");
  		}
  		System.out.println(res);
  	}
  
  	private static boolean is(int[] a, int[] b) {
  		for (int i = 0; i < a.length; i++)
  			if (a[i] != b[i])
  				return false;
  		return true;
  	}
  
  	private static long getSum(int st, int end, int seg, int l, int r) {
  		if (st >= l && r >= end) {
  			return sum[seg];
  		}
  		if (l > st || end > r) {
  			return 0;
  		}
  		int mid = (l + r) / 2;
  		return getSum(st, mid, 2 * seg + 1, l, r) + getSum(mid + 1, end, 2 * seg + 2, l, r);
  	}
  
  	private static void set(long diff, int l, int r, int ind, int seg) {
  		if (ind < l || ind > r)
  			return;
  		sum[seg] = sum[seg] + diff;
  		if (l != r) {
  			int mid = (l + r) / 2;
  			set(diff, l, mid, ind, 2 * seg + 1);
  			set(diff, mid + 1, r, ind, 2 * seg + 2);
  		}
  	}
  
  	private static long build(int seg, int l, int r) {
  		if (l == r) {
  			sum[seg] = a[l];
  			return a[l];
  		}
  		int m = (l + r) / 2;
  		sum[seg] = build(2 * seg + 1, l, m) + build(2 * seg + 2, m + 1, r);
  		return sum[seg];
  	}
  	
  }