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.*;

public class Random {

	static int MOD = 1000000007;
	static Scanner scn = new Scanner(System.in);

	public static void main(String[] args) {

		int t = scn.nextInt();
		while (t-- > 0) {
			int n = scn.nextInt();
			int s = scn.nextInt();

			int lo = 1;
			int hi = s / 2;

			if (n == 1) {
				System.out.println(s);
				continue;
			}

			int ans = 0;
			while (lo <= hi) {

				int mid = (lo + hi) / 2;

				if ((mid * ((n / 2) + 1)) <= s) {
					lo = mid + 1;
					ans = mid;
				} else {

					hi = mid - 1;
				}
			}
			System.out.println(ans);

		}

	}

	public static void printArray(int[] arr) {
		for (int i = 0; i < arr.length; i++) {
			System.out.print(arr[i] + " ");
		}
		System.out.println();
	}
}
