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.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashSet;

public class Main {
    public static void main(String[] args) throws java.lang.Exception {
        InputStreamReader in = new InputStreamReader(System.in);
        BufferedReader br = new BufferedReader(in);
        String s = br.readLine();
        long t = Long.parseLong(s);
//        int r=1;
        while (t-->0) {
             s = br.readLine();
             String ss[] = s.trim().split(" ");
             long n = Long.parseLong(ss[0]);
            long sum = Long.parseLong(ss[1]);
            if(n==1){
                System.out.println(sum);
                continue;
            }else if(n>sum){
                System.out.println(0);
                continue;
            }else if(n==sum){
                System.out.println(1);
                continue;
            }
            double x = (double) n/2;
            long y = n/2;
            long ans = 0;
            if(x==y){
            ans=sum/(n-y+1);
            }else {
                ans = sum/(n-(y+1)+1);
            }
            System.out.println(ans);

        }
    }
}