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: // package com.codeforces.global.round16.a;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  
  public class Solution {
      public static void main(String[] args) {
          FastScanner fs = new FastScanner();
          int t = fs.nextInt();
          while (t-- > 0) {
              long n = fs.nextLong(), s = fs.nextLong();
              if (n == 1) {
                  System.out.println(s);
              } else if (s == 1) {
                  System.out.println(0);
              } else {
                  // Idea. Use Greedy approach, consider the lower half of all n elements to be 0
                  // 0, 0, ... M, x_1, x_2, x_3, ... x_k s.t
                  // M + x_1 + x_2 + x_3 + ...x_k = s, this sequence will be of length
                  // Len = n / 2 + 1
                  // and each element will be s / Len
                  // Now s % Len != 0, we add the remaining sum to the last element
                  // x_k = s / Len + s % Len (why only last element ? Because to calculate median
                  // sequence must be sorted (non-decreasing). Hence rightmost element will be the greatest.)
                  long len = n / 2 + 1;
                  System.out.println(s / len);
              }
          }
  
      }
  
      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());
          }
      }
  }
  