#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a < 0) a = -a;
  if (b < 0) b = -b;
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
void read(T &res) {
  bool flag = false;
  char ch;
  while (!isdigit(ch = getchar())) (ch == '-') && (flag = true);
  for (res = ch - 48; isdigit(ch = getchar());
       res = (res << 1) + (res << 3) + ch - 48)
    ;
  flag && (res = -res);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long qp(long long a, long long b, long long mod) {
  long long ans = 1;
  if (b == 0) {
    return ans % mod;
  }
  while (b) {
    if (b % 2 == 1) {
      b--;
      ans = ans * a % mod;
    }
    a = a * a % mod;
    b = b / 2;
  }
  return ans % mod;
}
long long qpn(long long a, long long b, long long p) {
  long long ans = 1;
  a %= p;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % p;
      --b;
    }
    a = (a * a) % p;
    b >>= 1;
  }
  return ans % p;
}
signed main() {
  long long t;
  read(t);
  while (t--) {
    long long n, s;
    read(n);
    read(s);
    long long ans = s / n;
    ans = s / (n / 2 + 1);
    printf("%lld\n", ans);
  }
}

### 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;
long long gcd(long long a, long long b) {
  if (a < 0) a = -a;
  if (b < 0) b = -b;
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
void read(T &res) {
  bool flag = false;
  char ch;
  while (!isdigit(ch = getchar())) (ch == '-') && (flag = true);
  for (res = ch - 48; isdigit(ch = getchar());
       res = (res << 1) + (res << 3) + ch - 48)
    ;
  flag && (res = -res);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long qp(long long a, long long b, long long mod) {
  long long ans = 1;
  if (b == 0) {
    return ans % mod;
  }
  while (b) {
    if (b % 2 == 1) {
      b--;
      ans = ans * a % mod;
    }
    a = a * a % mod;
    b = b / 2;
  }
  return ans % mod;
}
long long qpn(long long a, long long b, long long p) {
  long long ans = 1;
  a %= p;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % p;
      --b;
    }
    a = (a * a) % p;
    b >>= 1;
  }
  return ans % p;
}
signed main() {
  long long t;
  read(t);
  while (t--) {
    long long n, s;
    read(n);
    read(s);
    long long ans = s / n;
    ans = s / (n / 2 + 1);
    printf("%lld\n", ans);
  }
}

```