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] 
/*

Sample Input 1

8
1 5
2 5
3 5
2 1
7 17
4 14
1 1000000000
1000000000 1

Sample Output 1

5
2
2
0
4
4
1000000000
0

*/
import java.io.*;
import java.util.*;
  
public class R1566A
{
    static int MAX(int x, int y){return ((x > y) ? x : y );}
    static int MIN(int x, int y){return ((x < y) ? x : y );}
    static int sum(int[] A){int sum = 0;for(int i=0; i<A.length; i++)sum+=A[i];return (sum);}
    static int gcd(int a, int b){if (b == 0)return a;return gcd(b, a % b);}
    static boolean isPowerOfTwo(int n){if(n==0)return false;return (int)(Math.ceil((Math.log(n) / Math.log(2)))) ==(int)(Math.floor(((Math.log(n) / Math.log(2)))));}
    static int countDigits(long l)
    {
            if (l >= 1000000000000000000L) return 19;
            if (l >= 100000000000000000L) return 18;
            if (l >= 10000000000000000L) return 17;
            if (l >= 1000000000000000L) return 16;
            if (l >= 100000000000000L) return 15;
            if (l >= 10000000000000L) return 14;
            if (l >= 1000000000000L) return 13;
            if (l >= 100000000000L) return 12;
            if (l >= 10000000000L) return 11;
            if (l >= 1000000000L) return 10;
            if (l >= 100000000L) return 9;
            if (l >= 10000000L) return 8;
            if (l >= 1000000L) return 7;
            if (l >= 100000L) return 6;
            if (l >= 10000L) return 5;
            if (l >= 1000L) return 4;
            if (l >= 100L) return 3;
            if (l >= 10L) return 2;
            return 1;
    }
    static void sort2D(int[][] Arr){Arrays.sort(Arr, (a, b) -> Double.compare(a[0], b[0]));}
    public static void main(String[] args) throws IOException
    {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        if (System.getProperty("ONLINE_JUDGE") == null)
        {
            try 
            {
                in = new BufferedReader(new FileReader("input.txt"));
                System.setOut(new PrintStream(new FileOutputStream("output.txt")));
            }
            catch (Exception e) {
                //System.out.println("Exception");
            }
        }
        int T = Integer.parseInt(in.readLine());
        while( T -- > 0)
        {
            String lines = in.readLine();
            String[] strs = lines.trim().split("\\s+");
            int n = Integer.parseInt(strs[0]);
            int s = Integer.parseInt(strs[1]);
            int temp = 1;
            if(s == 1)
                System.out.println("0");
            else if(n == 1)
                System.out.println(s);
            else
            {
                if(n % 2 == 0)
                    temp = n / 2 + 1;
                else
                    temp = (int)Math.ceil(n/(2*1.0));
                temp = s/temp;
                System.out.println(temp);
            }
        }
    }
}