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<vector>
#include<cmath>
#include<algorithm>
#include<string>
#include<utility>
#include<array>
#include<set>
#include <numeric>
#include <stdio.h>
#include <map>
#include <iomanip>
#include <sstream>
//#include <boost/math/common_factor_rt.hpp>
#include<fstream>
#define EPS (1e-7)
using namespace std;
#define ll long long



class Employee {
public:
	int id;
	int parent;
	vector<int> children;
	int salary;

	Employee() {
		
	}

	Employee(int num, int nparent) {
		id = num;
		parent = nparent;
		salary = -1;
	}



};



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



int main() {
	int A, B;
	cin >> A >> B;
	int ans;
	for (int i = 0; i < B; i++)
	{
		int kuti = i * A - (i - 1);
		if (kuti >= B) {
			ans = i;
			break;
		}
	}

	cout << ans << endl;

	return 0;
}