#include <bits/stdc++.h>
using namespace std;
template <typename Type>
void RIP(Type x) {
  cout << x << "\n";
}
template <typename Type, typename... Targs>
void RIP(Type x, Targs... args) {
  cout << x << " ";
  RIP(args...);
}
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const int N = 1e6 + 10;
const double eps = 1e-9;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mo = (1 ? 998244353 : 1000000007);
long long mul(long long a, long long b, long long m = mo) { return a * b % m; }
long long fpow(long long a, long long b, long long m = mo) {
  long long ans = 1;
  for (; b; a = mul(a, a, m), b >>= 1)
    if (b & 1) ans = mul(ans, a, m);
  return ans;
}
inline long long read() {
  long long x = 0, tag = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') tag = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - 48;
  return x * tag;
}
const double pi = acos(-1.0);
double readf() {
  double x;
  if (scanf("%lf", &x) != 1) exit(0);
  return x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
long long a[N];
void Solve() {
  long long n = read(), m = read();
  cout << m / (n + 1 - (n + 1 >> 1)) << "\n";
}
int main() {
  int T = 1;
  T = read();
  for (int kase = (1); kase <= (T); ++kase) {
    Solve();
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 Type>
void RIP(Type x) {
  cout << x << "\n";
}
template <typename Type, typename... Targs>
void RIP(Type x, Targs... args) {
  cout << x << " ";
  RIP(args...);
}
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const int N = 1e6 + 10;
const double eps = 1e-9;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mo = (1 ? 998244353 : 1000000007);
long long mul(long long a, long long b, long long m = mo) { return a * b % m; }
long long fpow(long long a, long long b, long long m = mo) {
  long long ans = 1;
  for (; b; a = mul(a, a, m), b >>= 1)
    if (b & 1) ans = mul(ans, a, m);
  return ans;
}
inline long long read() {
  long long x = 0, tag = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') tag = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - 48;
  return x * tag;
}
const double pi = acos(-1.0);
double readf() {
  double x;
  if (scanf("%lf", &x) != 1) exit(0);
  return x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
long long a[N];
void Solve() {
  long long n = read(), m = read();
  cout << m / (n + 1 - (n + 1 >> 1)) << "\n";
}
int main() {
  int T = 1;
  T = read();
  for (int kase = (1); kase <= (T); ++kase) {
    Solve();
  }
  return 0;
}

```