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 queue
from math import *
import sys
from collections import *
from random import *
from itertools import *

sys.setrecursionlimit(99999)
eps = sys.float_info.epsilon
P = 2
INF = 1e9 + 1
MOD = 1000000007


def is_prime(n):
    if n == 0 or n == 1:
        return False
    d = 2
    while d * d <= n:
        if n % d == 0:
            return False
        d += 1
    return True


def div_up(n, k):
    if n % k == 0:
        return n // k
    else:
        return n // k + 1


def num_len(n, base = 10):
    if n == 0:
        return 1
    return int(floor(log(n, base) + 1))


def dfs_sample(graph, cl, p, v):
    cl[v] = 1
    for to in graph[v]:
        if cl[to] == 1 and p[v] != to:
            # yay, cycle
            pass
        elif cl[to] == 0:
            p[to] = v
            dfs_sample(graph, cl, p, to)
    cl[v] = 2


def down(a):
    for i in range(1, len(a)):
        if a[i] > a[i - 1]:
            return False
    return True


def up(a):
    for i in range(1, len(a)):
        if a[i] < a[i - 1]:
            return False
    return True


def code(c):
    return ord(c) - 32


def _hash_(s):
    res, p = 0, 1
    for i in range(len(s)):
        res += (code(s[i]) * p)
        res %= MOD
        p *= P
        p %= MOD
    return res % MOD


def remove_edge(v, u, graph):
    graph[v].remove(u)
    graph[u].remove(v)


b = True
def dfs_stack(start, graph, visited, flags):
    global b
    stack = []
    visited[start] = True
    stack.append(start)
    while len(stack) != 0:
        v = stack.pop()
        if not flags[v]:
            b = False
        for to in graph[v]:
            if not visited[to]:
                visited[to] = True
                stack.append(to)


def all_eq(a):
    for i in range(len(a) - 1):
        if a[i] != a[i + 1]:
            return False
    return True


def solve():
    n, s = map(int, input().split())
    if n == 1:
        print(s)
    else:
        if n % 2 != 0:
            print(2 * s // (n + 1))
        else:
            print(2 * s // (n + 2))


for _ in range(int(input())):
    solve()


def debug():
    pass


# debug()