#include <bits/stdc++.h>
using namespace std;
bool isPrime(long long n) {
  if (n == 1) return false;
  if (n == 2) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
const long long N = 1e7 + 5;
bool sieve[N];
void soe() {
  for (long long i = 2; i < N; ++i) sieve[i] = true;
  for (long long i = 2; i * i < N; ++i) {
    if (sieve[i] == true) {
      for (long long j = i * i; j < N; j += i) {
        sieve[j] = false;
      }
    }
  }
}
long long countDivisors(long long n) {
  long long cnt = 0;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return cnt;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fact(long long n);
long long nCr(long long n, long long r) {
  return fact(n) / (fact(r) * fact(n - r));
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) res = res * i;
  return res;
}
long long PowMod(long long n) {
  long long ret = 1;
  long long a = 2;
  while (n > 0) {
    if (n & 1) ret = ret * a % 1000000007;
    a = a * a % 1000000007;
    n >>= 1;
  }
  return ret;
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  ;
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long div = 1 + n / 2;
    cout << s / div << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
bool isPrime(long long n) {
  if (n == 1) return false;
  if (n == 2) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
const long long N = 1e7 + 5;
bool sieve[N];
void soe() {
  for (long long i = 2; i < N; ++i) sieve[i] = true;
  for (long long i = 2; i * i < N; ++i) {
    if (sieve[i] == true) {
      for (long long j = i * i; j < N; j += i) {
        sieve[j] = false;
      }
    }
  }
}
long long countDivisors(long long n) {
  long long cnt = 0;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return cnt;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fact(long long n);
long long nCr(long long n, long long r) {
  return fact(n) / (fact(r) * fact(n - r));
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) res = res * i;
  return res;
}
long long PowMod(long long n) {
  long long ret = 1;
  long long a = 2;
  while (n > 0) {
    if (n & 1) ret = ret * a % 1000000007;
    a = a * a % 1000000007;
    n >>= 1;
  }
  return ret;
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  ;
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long div = 1 + n / 2;
    cout << s / div << endl;
  }
  return 0;
}

```