#include <bits/stdc++.h>
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
std::pair<int, int> operator+(std::pair<int, int> a, std::pair<int, int> b) {
  return {a.first + b.first, a.second + b.second};
}
std::pair<long long, long long> operator+(std::pair<long long, long long> a,
                                          std::pair<long long, long long> b) {
  return {a.first + b.first, a.second + b.second};
}
std::ostream& operator<<(std::ostream& out, std::pair<int, int> a) {
  out << a.first << " " << a.second << "\n";
  return out;
}
std::ostream& operator<<(std::ostream& out, std::pair<long long, long long> a) {
  out << a.first << " " << a.second << "\n";
  return out;
}
std::istream& operator>>(std::istream& in, std::pair<int, int>& a) {
  in >> a.first >> a.second;
  return in;
}
std::istream& operator>>(std::istream& in, std::pair<long long, long long>& a) {
  in >> a.first >> a.second;
  return in;
}
long long bpow(long long x, long long v) {
  if (v == 0) return 1 % 1000000007;
  if (v % 2 == 1) return (x * bpow(x, v - 1)) % 1000000007;
  long long r = bpow(x, v / 2);
  return (r * r) % 1000000007;
}
using namespace std;
void solve() {
  long long n, s;
  cin >> n >> s;
  long long a = n - n / 2 + (1 - n % 2);
  cout << s / a << "\n";
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

### Prompt

In cpp, your task is to solve the following problem:

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] 

### Response

```cpp
#include <bits/stdc++.h>
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
std::pair<int, int> operator+(std::pair<int, int> a, std::pair<int, int> b) {
  return {a.first + b.first, a.second + b.second};
}
std::pair<long long, long long> operator+(std::pair<long long, long long> a,
                                          std::pair<long long, long long> b) {
  return {a.first + b.first, a.second + b.second};
}
std::ostream& operator<<(std::ostream& out, std::pair<int, int> a) {
  out << a.first << " " << a.second << "\n";
  return out;
}
std::ostream& operator<<(std::ostream& out, std::pair<long long, long long> a) {
  out << a.first << " " << a.second << "\n";
  return out;
}
std::istream& operator>>(std::istream& in, std::pair<int, int>& a) {
  in >> a.first >> a.second;
  return in;
}
std::istream& operator>>(std::istream& in, std::pair<long long, long long>& a) {
  in >> a.first >> a.second;
  return in;
}
long long bpow(long long x, long long v) {
  if (v == 0) return 1 % 1000000007;
  if (v % 2 == 1) return (x * bpow(x, v - 1)) % 1000000007;
  long long r = bpow(x, v / 2);
  return (r * r) % 1000000007;
}
using namespace std;
void solve() {
  long long n, s;
  cin >> n >> s;
  long long a = n - n / 2 + (1 - n % 2);
  cout << s / a << "\n";
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

```