#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
T min_(T a, T b) {
  return a > b ? b : a;
}
template <typename T, typename... Ts>
T min_(T first, Ts... last) {
  return min_(first, min_(last...));
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
template <class T, class S>
std::ostream& operator<<(std::ostream& os, const std::pair<T, S>& t) {
  os << "(" << t.first << ", " << t.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  out << "[";
  size_t last = v.size() - 1;
  for (size_t i = 0; i < v.size(); ++i) {
    out << v[i];
    if (i != last) out << ", ";
  }
  out << "]";
  return out;
}
void solve() {
  int n, s;
  cin >> n >> s;
  n -= (n + 1) / 2 - 1;
  cout << s / n << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
T min_(T a, T b) {
  return a > b ? b : a;
}
template <typename T, typename... Ts>
T min_(T first, Ts... last) {
  return min_(first, min_(last...));
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
template <class T, class S>
std::ostream& operator<<(std::ostream& os, const std::pair<T, S>& t) {
  os << "(" << t.first << ", " << t.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  out << "[";
  size_t last = v.size() - 1;
  for (size_t i = 0; i < v.size(); ++i) {
    out << v[i];
    if (i != last) out << ", ";
  }
  out << "]";
  return out;
}
void solve() {
  int n, s;
  cin >> n >> s;
  n -= (n + 1) / 2 - 1;
  cout << s / n << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}

```