#include <bits/stdc++.h>
using namespace std;
template <typename T>
void error(T it) {}
template <typename Ty, typename T, typename... Args>
void error(Ty it, T a, Args... arg) {
  cout << *it << " = " << a << '\t';
  error((++it), arg...);
}
void _split(string s, vector<string> &_os) {
  string se;
  long long x = 0;
  for (auto i : s) {
    if (i == ',' && x == 0)
      _os.push_back(se), se = "";
    else
      se += i;
    if (i == '(') x++;
    if (i == ')') x--;
  }
  _os.push_back(se);
}
long long mod = 1e+9 + 7;
long long inf = (2e+18);
long double pi = 3.1415926536;
const long long N = 2e+5 + 3;
const long long _N = 2e+6 + 3;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  cin >> test;
  while (test--) {
    long long n, s;
    cin >> n >> s;
    long long m;
    if (n % 2 == 0) {
      m = n / 2 + 1;
    } else {
      m = (n + 1) / 2;
    }
    long long t1 = s % m;
    long long t2 = (s - t1) / m;
    cout << t2 << endl;
  }
  cerr << "\nTime : "
       << 1000 * (long double)clock() / (long double)CLOCKS_PER_SEC << "ms\t";
}

### Prompt

Please create a solution in Cpp to 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>
using namespace std;
template <typename T>
void error(T it) {}
template <typename Ty, typename T, typename... Args>
void error(Ty it, T a, Args... arg) {
  cout << *it << " = " << a << '\t';
  error((++it), arg...);
}
void _split(string s, vector<string> &_os) {
  string se;
  long long x = 0;
  for (auto i : s) {
    if (i == ',' && x == 0)
      _os.push_back(se), se = "";
    else
      se += i;
    if (i == '(') x++;
    if (i == ')') x--;
  }
  _os.push_back(se);
}
long long mod = 1e+9 + 7;
long long inf = (2e+18);
long double pi = 3.1415926536;
const long long N = 2e+5 + 3;
const long long _N = 2e+6 + 3;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  cin >> test;
  while (test--) {
    long long n, s;
    cin >> n >> s;
    long long m;
    if (n % 2 == 0) {
      m = n / 2 + 1;
    } else {
      m = (n + 1) / 2;
    }
    long long t1 = s % m;
    long long t2 = (s - t1) / m;
    cout << t2 << endl;
  }
  cerr << "\nTime : "
       << 1000 * (long double)clock() / (long double)CLOCKS_PER_SEC << "ms\t";
}

```