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.util.*;
import java.io.*;
public class Main {
	public static void main(String[] args) throws IOException, InterruptedException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int jj = sc.nextInt();
		while(jj-->0) {
			long n = sc.nextLong();
			long s = sc.nextLong();
			
				long div  = (n/2)+1;
				if(n == 1) {
					pw.println(s);
				}
				else {
					if(s%div==0) {
						
						pw.println(s/div);
					}
					else {
						//pw.println("M");
						pw.println(s/div);
					}
				}
				
				
			
		}
				
		
		pw.flush();
	}
	
	
}

class edge implements Comparable<edge> {
	int from;
	int dest;
	double cost;

	public edge(int from, int dest, double cost) {
		this.from = from;
		this.dest = dest;
		this.cost = cost;

	}
public String toString() {
	return from+" "+dest+" "+cost;
}
	@Override
	public int compareTo(edge o) {
//		return cost - o.cost;
		if (cost < o.cost)
			return -1;
		if (cost > o.cost)
			return 1;
		return 0;
	}

}

class pair {
	int a;
	int b;

	public pair(int a, int b) {
		this.a = a;
		this.b = b;
	}

}

class Scanner {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s) {
		br = new BufferedReader(new InputStreamReader(s));
	}

	public String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	public long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	public String nextLine() throws IOException {
		return br.readLine();
	}

	public double nextDouble() throws IOException {
		String x = next();
		StringBuilder sb = new StringBuilder("0");
		double res = 0, f = 1;
		boolean dec = false, neg = false;
		int start = 0;
		if (x.charAt(0) == '-') {
			neg = true;
			start++;
		}
		for (int i = start; i < x.length(); i++)
			if (x.charAt(i) == '.') {
				res = Long.parseLong(sb.toString());
				sb = new StringBuilder("0");
				dec = true;
			} else {
				sb.append(x.charAt(i));
				if (dec)
					f *= 10;
			}
		res += Long.parseLong(sb.toString()) / f;
		return res * (neg ? -1 : 1);
	}

	public boolean ready() throws IOException {
		return br.ready();
	}
}