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] 
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
long long fastPower(long long base, long long power) {
  long long res = 1;
  while (power) {
    if (power & 1) res *= base;
    base *= base;
    power /= 2;
  }
  return res;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long isp(long long n) {
  if (n == 1) {
    return 0;
  }
  if (n == 2) {
    return 1;
  }
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      return 0;
    }
  }
  return 1;
}
bool good[N];
bool seive() {
  memset(good, true, sizeof good);
  good[1] = false;
  for (long long i = 2; i < N; i++) {
    if (good[i]) {
      for (long long j = 1ll * i * i; j < N; j += i) {
        good[j] = false;
      }
    }
  }
}
int commDiv(int a, int b) {
  int n = gcd(a, b);
  int result = 0;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        result += 1;
      else
        result += 2;
    }
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long ans = (n + 1) / 2;
    cout << s / (n - ans + 1) << endl;
  }
}
