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] 
#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (int i = 0; i < v.size(); ++i) {
    os << v[i];
    if (i != v.size() - 1) os << ", ";
  }
  os << "]\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const deque<T>& v) {
  os << "[";
  for (int i = 0; i < v.size(); ++i) {
    os << v[i];
    if (i != v.size() - 1) os << ", ";
  }
  os << "]\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  for (auto it : v) {
    os << it;
    if (it != *v.rbegin()) os << ", ";
  }
  os << "]\n";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const map<T, S>& v) {
  for (auto it : v) os << it.first << " : " << it.second << "\n";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  os << "(";
  os << v.first << ", " << v.second << ")";
  return os;
}
void go() { cin.tie(nullptr)->sync_with_stdio(false); }
void solve_case() {
  long long n, s, d;
  cin >> n >> s;
  if (n % 2 == 0) {
    d = n / 2 + 1;
  } else {
    d = (n + 1) / 2;
  }
  long long ans = s / d;
  cout << ans << '\n';
}
signed main() {
  go();
  long long t;
  cin >> t;
  while (t--) {
    solve_case();
  }
}
