#include <bits/stdc++.h>
using namespace std;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2> &p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
long long int n, m;
vector<long long int> adj[1];
bool visited[1];
long long int start[1];
long long int end[1];
long long int parent[1];
long long int level[1];
vector<long long int> dfs_order;
void dfs(long long int node) {
  visited[node] = true;
  for (long long int next : adj[node]) {
    if (!visited[next]) {
      parent[next] = node;
      level[next] = level[node] + 1;
      dfs(next);
    }
  }
}
long long int dist[1];
void bfs(long long int start) {
  memset(dist, -1, sizeof dist);
  queue<long long int> q;
  dist[start] = 0;
  q.push(start);
  while (!q.empty()) {
    long long int v = q.front();
    q.pop();
    for (long long int e : adj[v]) {
      if (dist[e] == -1) {
        dist[e] = dist[v] + 1;
        q.push(e);
      }
    }
  }
}
long long int lift(long long int a, long long int dist,
                   vector<vector<long long int>> &up) {
  for (long long int i = 0; i < 20; i++) {
    if (dist & (1 << i)) {
      a = up[a][i];
    }
  }
  return a;
}
void preprocesslift(vector<vector<long long int>> &up) {
  for (long long int j = 1; j < 20; j++) {
    for (long long int i = 1; i <= n; i++) {
      if (up[i][j - 1] != -1) {
        up[i][j] = up[up[i][j - 1]][j - 1];
      }
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t > 0) {
    long long int n, s;
    cin >> n >> s;
    if (((n / 2) + 1) > s)
      cout << "0\n";
    else if (s == 1)
      cout << n << "\n";
    else {
      long long int ans = 0;
      if (n % 2 == 0)
        ans = s / ((n / 2) + 1);
      else
        ans = s / ((n / 2) + 1);
      cout << ans << "\n";
    }
    t--;
  }
  return 0;
}

### Prompt

Generate 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;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2> &p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
long long int n, m;
vector<long long int> adj[1];
bool visited[1];
long long int start[1];
long long int end[1];
long long int parent[1];
long long int level[1];
vector<long long int> dfs_order;
void dfs(long long int node) {
  visited[node] = true;
  for (long long int next : adj[node]) {
    if (!visited[next]) {
      parent[next] = node;
      level[next] = level[node] + 1;
      dfs(next);
    }
  }
}
long long int dist[1];
void bfs(long long int start) {
  memset(dist, -1, sizeof dist);
  queue<long long int> q;
  dist[start] = 0;
  q.push(start);
  while (!q.empty()) {
    long long int v = q.front();
    q.pop();
    for (long long int e : adj[v]) {
      if (dist[e] == -1) {
        dist[e] = dist[v] + 1;
        q.push(e);
      }
    }
  }
}
long long int lift(long long int a, long long int dist,
                   vector<vector<long long int>> &up) {
  for (long long int i = 0; i < 20; i++) {
    if (dist & (1 << i)) {
      a = up[a][i];
    }
  }
  return a;
}
void preprocesslift(vector<vector<long long int>> &up) {
  for (long long int j = 1; j < 20; j++) {
    for (long long int i = 1; i <= n; i++) {
      if (up[i][j - 1] != -1) {
        up[i][j] = up[up[i][j - 1]][j - 1];
      }
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t > 0) {
    long long int n, s;
    cin >> n >> s;
    if (((n / 2) + 1) > s)
      cout << "0\n";
    else if (s == 1)
      cout << n << "\n";
    else {
      long long int ans = 0;
      if (n % 2 == 0)
        ans = s / ((n / 2) + 1);
      else
        ans = s / ((n / 2) + 1);
      cout << ans << "\n";
    }
    t--;
  }
  return 0;
}

```