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 long mod = 1000000007;
  		static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  		public static void main(String[] args) throws IOException  {
  			FastReader sc = new FastReader();
  			int t  = sc.nextInt();
  			while( t-- > 0) {
  				int n = sc.nextInt();
  				int mid = (n+1)/2;
  				mid = n - mid;
  				long sum = sc.nextLong();
  				// (x+1)(n-mid) + x = sum
  				out.println((sum)/(mid+1));
  			}
  			out.flush();
  		}		
  
  		public static int[] nextLargerElement(int[] arr, int n)	{ 
  			Stack<Integer> stack = new Stack<>();
  			int rtrn[] = new int[n];
  	        rtrn[n-1] = -1;
  	        stack.push( n-1);
  	        for( int i = n-2 ;i >= 0 ; i--){
  	            int temp = arr[i];
  	            int lol = -1;
  	            while( !stack.isEmpty() && arr[stack.peek()] <= temp){
  	            	if(arr[stack.peek()] == temp ) {
  	            		lol = stack.peek();
  	            	}
  	                stack.pop();
  	            }
  	            if( stack.isEmpty()){
  	            	if( lol != -1) {
  	            		rtrn[i] = lol;
  	            		}
  	            	else {
  	                rtrn[i] = -1;
  	            	}
  	            }
  	            else{
  	            	rtrn[i] = stack.peek();
  	            }
  	            stack.push( i);
  	        }
  	        return rtrn;
  		}
  		
  		 static int gcd(int a, int b)
  		    {
  		        if (a == 0)
  		            return b;
  		        return gcd(b % a, a);
  		    }
  		     
  //		    // method to return LCM of two numbers
  //		    static int lcm(int a, int b)
  //		    {
  //		        return (a / gcd(a, b)) * b;
  //		    }
  	 
  		static HashMap<Long,Long> primefactor( long n){
  			HashMap<Long ,Long> hm = new HashMap<>();
  			long temp = 0;
  			while( n%2 == 0) {
  				temp++;
  				n/=2;
  	 		 }	
  			if( temp!= 0) {
  	 			 hm.put( 2L, temp);
  	 		 }
  	 		 long c = (long)Math.sqrt(n);
  	 		 for( long i = 3 ; i <= c ; i+=2) {
  	 			 temp = 0;
  	 			 while( n% i == 0) {
  	 				 temp++;
  	 				 n/=i;
  	 			 }
  	 			 if( temp!= 0) {
  	 				 hm.put( i, temp);
  	 			 }
  	 		 }
  	 		 if( n!= 1) {
  	 			 hm.put( n , 1L);
  	 		 }
  	 		 return hm;
  	 		// this is the link where i found this method
  	 		 // https://www.geeksforgeeks.org/print-all-prime-factors-of-a-given-number/
  	 	 }
  	 
  		 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;
  		        }
  		    }
  		
  }