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] 
import java.io.PrintWriter;
import java.util.*;
public class Main
{
    public static void main(String[] args) {
        Scanner obj=new Scanner(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int length=obj.nextInt();
        while(length--!=0) {
            int n = obj.nextInt();
            int s = obj.nextInt();
            int each = s / n;
            int mid = ceil(n, 2);
            int extra = (mid - 1) * each + s%n;
            int imp = n - mid + 1;
            int ans = each + (extra / imp);
            if (extra > imp && extra%imp>=imp)ans++;
            out.println(ans);
        }
        out.flush();
    }
    public static int ceil(int a,int b)
    {
        return (a+b-1)/b;
    }
}