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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <numeric>
#include <queue>
#include <unordered_map>
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <sstream>
#include <stack>
#include <map>
#include <set>
#include <ios>
#include <cctype>
#include <cstdio>
#include <functional>
#include <cassert>
#include <fstream>
#include <ctime>

using namespace std;
using ll = long long;

#define REP(i,a) for(int i = 0;i < (a);++i)
#define FOR(i,a,b) for(int i = (a);i < (b); ++i)
#define FORR(i,a,b) for(int i = (a) - 1;i >=(b);--i)
#define ALL(obj) (obj).begin(),(obj).end()
#define SIZE(obj) (int)(obj).sizeT()
#define YESNO(cond,yes,no){cout <<((cond)?(yes):(no))<<endl; }
#define SORT(list) sort(ALL((list)));
#define RSORT(list) sort((list).rbegin(),(list).rend())
#define ASSERT(cond,mes) assert(cond && mes)

constexpr int MOD = 1'000'000'007;
constexpr int INF = 1'050'000'000;

template<typename T>
T round_up(const T& a, const T& b) {
    return (a + (b - 1)) / b;
}

template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& p) {
    is >> p.first >> p.second;
    return is;
}

template <typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2>& p) {
    os << p.first << p.second;
    return os;
}

template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
    REP(i, (int)v.size())is >> v[i];
    return is;
}

template <typename T>
T clamp(T& n, T a, T b) {
    if (n < a)n = a;
    if (n > b)n = b;
    return n;
}

template <typename T>
static T GCD(T u, T v) {
    T r;
    while (v != 0) {
        r = u % v;
        u = v;
        v = r;
    }
    return u;
}

template <typename T>
static T LCM(T u, T v) {
    return u / GCD(u, v) * v;
}

std::vector<int> enum_div(int n) {
    std::vector<int> ret;
    for (int i = 1; i*i <= n; ++i) {
        if (n%i == 0) {
            ret.push_back(i);
            if (i != 1 && i*i != n) {
                ret.push_back(n / i);
            }
        }
    }
    return ret;
}

template<typename T>
bool chmin(T& a, const T& b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template<typename T>
bool chmax(T& a, const T& b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

struct ToUpper {
    char operator()(char c) { return toupper(c); }
};

struct ToLower {
    char operator()(char c) { return tolower(c); }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    int A, B;
    cin >> A >> B;

    int ans = 0;
    int num = 1;
    while (num < B) {
        num--;
        num += A;
        ans++;
    }

    cout << ans << endl;
    return 0;
}