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.*;
import java.util.*;

public class Main {
    public static void main(String args[]) {
        FastReader fs = new FastReader();
        int t = fs.nextInt();
        while(t-- >0){
            int n = fs.nextInt();
            int s = fs.nextInt();
            if(n%2==0) System.out.println(s / (n -n/2 + 1));
            else System.out.println(s / (n -n/2));
        }
    }
}

class FastReader{
    BufferedReader br;
    StringTokenizer st;

    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());
    }
    int[] nextArr(int n){
        int[] a = new int[n];
        for(int i=0; i<n; i++){
            a[i] = nextInt();
        }
        return a;
    }
}