#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
long long M = 1e9 + 7;
long long mod(long long x) { return ((x % M + M) % M); }
long long gcd(long long a, long long b);
long long lcm(long long a, long long b);
bool prime(long long n);
long long BinExp(long long base, long long power);
long long ModInverse(long long base);
long long nCr(long long n, long long r);
long long fact(long long n) { return (n <= 1) ? 1 : n * fact(n - 1); }
void solve() {
  long long n, s;
  cin >> n >> s;
  n -= (n - 1) / 2;
  cout << (long long)(s / n) << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
}
long long nCr(long long n, long long r) {
  r = (n - r <= r) ? n - r : r;
  long long ans = 1;
  for (long long i = 0; i < r; i++) {
    ans *= (n - i);
    ans /= (i + 1);
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return (a == 1) ? a : gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
bool prime(long long n) {
  if (n < 2) return false;
  for (long long i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return false;
  return true;
}
long long BinExp(long long base, long long power) {
  if (!power) return 1;
  long long res = 1;
  while (power > 1) {
    if (power % 2) {
      power--;
      res *= base;
    } else {
      base *= base;
      power /= 2;
    }
  }
  return base * res;
}
long long ModInverse(long long base) {
  long long power = M - 2;
  if (!power) return 1;
  long long res = 1;
  while (power > 1) {
    if (power % 2) {
      power--;
      res = mod(res * base);
    } else {
      base = mod(base * base);
      power /= 2;
    }
  }
  return mod(base * res);
}

### 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
long long M = 1e9 + 7;
long long mod(long long x) { return ((x % M + M) % M); }
long long gcd(long long a, long long b);
long long lcm(long long a, long long b);
bool prime(long long n);
long long BinExp(long long base, long long power);
long long ModInverse(long long base);
long long nCr(long long n, long long r);
long long fact(long long n) { return (n <= 1) ? 1 : n * fact(n - 1); }
void solve() {
  long long n, s;
  cin >> n >> s;
  n -= (n - 1) / 2;
  cout << (long long)(s / n) << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
}
long long nCr(long long n, long long r) {
  r = (n - r <= r) ? n - r : r;
  long long ans = 1;
  for (long long i = 0; i < r; i++) {
    ans *= (n - i);
    ans /= (i + 1);
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return (a == 1) ? a : gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
bool prime(long long n) {
  if (n < 2) return false;
  for (long long i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return false;
  return true;
}
long long BinExp(long long base, long long power) {
  if (!power) return 1;
  long long res = 1;
  while (power > 1) {
    if (power % 2) {
      power--;
      res *= base;
    } else {
      base *= base;
      power /= 2;
    }
  }
  return base * res;
}
long long ModInverse(long long base) {
  long long power = M - 2;
  if (!power) return 1;
  long long res = 1;
  while (power > 1) {
    if (power % 2) {
      power--;
      res = mod(res * base);
    } else {
      base = mod(base * base);
      power /= 2;
    }
  }
  return mod(base * res);
}

```