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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Arrays;




public class Codeforces {
 public static void main(String[] args) throws IOException{

       BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
       BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
       
       long t = Long.parseLong(br.readLine().trim());
        
       while(t-->0){
             String arr[] = br.readLine().split(" ");
             long s = Long.parseLong(arr[1].trim());
             long n = Long.parseLong(arr[0].trim());
              long ans =0;
             if(n==1){System.out.println(s);
            }
            else if(s==1){
                System.out.println("0");
            }
            else{
              if(n%2==0){
                ans = (s/((n/2)+1));
              }
              else{
                  ans = (s/(n/2+1));
              }
              System.out.println(ans);
            }
            }
       br.close();
       log.flush();
   } 

static void print(int arr[]) {
    System.out.print("Print :- ");
    for(int i=0 ; i<arr.length ; i++) {
        System.out.print(arr[i]+" ");
    }
    System.out.println();
        }   // end-print


static boolean isPrime(long n){
    if (n <= 1)  return false;

    else if (n == 2) return true;

    else if (n % 2 == 0)return false;

    for (int i = 3; i <= Math.sqrt(n); i += 2){
        if (n % i == 0) return false;
    }
    return true;
        }//end isprime
 

static boolean isPrime(char a){
    if (a == '2' || a == '3' || a == '5' || a == '7') {
        return true;
    }

    return false;
    
}


static int computeXOR(int n){
        if (n % 4 == 0)
            return n;

        if (n % 4 == 1)
            return 1;
      
       
        if (n % 4 == 2)
            return n + 1;
      
        return 0;
        }
        // end compute Xor
}
