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] 
//package CompleteSearch.BruteForce;

import java.util.Scanner;

public class isithalloween 
{
    public static void main(String args[])
    {
        Scanner bob = new Scanner(System.in);
      
        int t = bob.nextInt();
        
        for(int i=0; i<t; i++)
        {
            double size = bob.nextInt();
            double sum = bob.nextInt();
            
            if(size == 1)
            {
                System.out.println((int) sum);
            }
            else if(sum == 1)
            {
                System.out.println(0);
            }
            else
            {
                if(size%2==0)
                {
                    if((size/2)%2==0)
                    {
                        
                        System.out.println( (int) Math.ceil(sum/size));
                    }
                    else
                    {
                        
                        System.out.println( (int) Math.floor( sum / size));
                    }
                }
                else
                {
                    System.out.println( (int) Math.floor( sum / ((size+1)/2)) );
                }
            }
            
            

        }
        
        
        
    }
}
