#include <bits/stdc++.h>
using namespace std;
const long long INF_LL = 1e18;
const int INF = 1e9 + 100;
const long double EPS = 1e-10;
const int BLOCK = 550;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
void open_file() {}
const int maxN = 2e5 + 100;
const int MOD = 1e9 + 7;
void sol() {
  int n, s;
  cin >> n >> s;
  int L = 1, R = s;
  int med = (n + 1) / 2;
  int ans = 0;
  while (L <= R) {
    int mid = (L + R) >> 1;
    long long sum = 1LL * (n - med + 1) * mid;
    if (sum <= s) {
      ans = mid;
      L = mid + 1;
    } else
      R = mid - 1;
  }
  cout << ans << '\n';
}
void solve() {
  int T = 1;
  cin >> T;
  int TestCase = 0;
  while (T--) {
    TestCase += 1;
    sol();
  }
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  open_file();
  solve();
  return 0;
}

### 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;
const long long INF_LL = 1e18;
const int INF = 1e9 + 100;
const long double EPS = 1e-10;
const int BLOCK = 550;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
void open_file() {}
const int maxN = 2e5 + 100;
const int MOD = 1e9 + 7;
void sol() {
  int n, s;
  cin >> n >> s;
  int L = 1, R = s;
  int med = (n + 1) / 2;
  int ans = 0;
  while (L <= R) {
    int mid = (L + R) >> 1;
    long long sum = 1LL * (n - med + 1) * mid;
    if (sum <= s) {
      ans = mid;
      L = mid + 1;
    } else
      R = mid - 1;
  }
  cout << ans << '\n';
}
void solve() {
  int T = 1;
  cin >> T;
  int TestCase = 0;
  while (T--) {
    TestCase += 1;
    sol();
  }
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  open_file();
  solve();
  return 0;
}

```