#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10, mod = 1e9 + 9;
const int maxn = 500010;
const long double eps = 1e-5;
const int EPS = 500 * 500;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long first = 1;
  while (!isdigit(ch)) {
    if (ch == '-') first *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= first;
}
template <typename T, typename... Args>
void read(T &first, Args &...args) {
  read(first);
  read(args...);
}
int main() {
  std::ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int _;
  cin >> _;
  while (_--) {
    int n, second;
    cin >> n >> second;
    cout << second / (n / 2 + 1) << '\n';
  }
}

### Prompt

Your task is to create a cpp solution 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;
const int N = 2e6 + 10, mod = 1e9 + 9;
const int maxn = 500010;
const long double eps = 1e-5;
const int EPS = 500 * 500;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long first = 1;
  while (!isdigit(ch)) {
    if (ch == '-') first *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= first;
}
template <typename T, typename... Args>
void read(T &first, Args &...args) {
  read(first);
  read(args...);
}
int main() {
  std::ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int _;
  cin >> _;
  while (_--) {
    int n, second;
    cin >> n >> second;
    cout << second / (n / 2 + 1) << '\n';
  }
}

```