Problem:
Score : 200 points

Problem Statement

Takahashi's house has only one socket.

Takahashi wants to extend it with some number of power strips, each with A sockets, into B or more empty sockets.

One power strip with A sockets can extend one empty socket into A empty sockets.

Find the minimum number of power strips required.

Constraints

All values in input are integers.

2 \leq A \leq 20

1 \leq B \leq 20

Input

Input is given from Standard Input in the following format:

A B

Output

Print the minimum number of power strips required.

Sample Input 1

4 10

Sample Output 1

3

3 power strips, each with 4 sockets, extend the socket into 10 empty sockets.

Sample Input 2

8 9

Sample Output 2

2

2 power strips, each with 8 sockets, extend the socket into 15 empty sockets.

Sample Input 3

8 8

Sample Output 3

1

Solution in C++ (Clang 10.0.0) (judge verdict: Wrong Answer)

//
#include <vector>
#include <set>
#include <map>
#include <array>
#include <iostream>
#include <cstdio>
#include <cstdint>
#include <algorithm>
#include <numeric>
#include <bitset>
//
using ll = long long;
using namespace std;
#define rep(i,n) for(ll i=0;i<(n);++i)
inline void yes(bool cond) { cond ? puts("Yes") : puts("No"); }
template<typename Type> inline void chmin(Type& a, Type b) { if (a > b) a = b; }
template<typename Type> inline void chmax(Type& a, Type b) { if (a < b) a = b; }
template<typename Arr, typename Type> inline void fill(Arr& arr, Type v) { std::fill(arr.begin(), arr.end(), v); }
template<typename Type> inline void sort(Type& arr) { std::sort(arr.begin(), arr.end()); }
template<typename Type> inline Type nth(vector<Type>& arr, size_t pos) { std::nth_element(arr.begin(), arr.begin() + pos, arr.end()); return arr[pos]; }
constexpr double BIGP = 1000000007L;
constexpr double INF_I = std::numeric_limits<double>::max();
constexpr double INF_F = std::numeric_limits<float>::infinity();
constexpr double PI = 3.14159265358979323846L;
//
int main()
{
    cin.tie(0); ios::sync_with_stdio(false);
    ll A, B;
    cin >> A >> B;

    int total = +1;
    int n = 0;
    for (n = 1; n < 100; ++n)
    {
        total -= 1;
        total += A;
        if (total >= B)
        {
            break;
        }
    }

    cout << n;
    return 0;
}