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 <bits/stdc++.h>
#include<iostream>
#include<queue>
using namespace std;
#define ll long long
#define PI 3.1415926535898
/*
int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n; ll n1[100005];
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> n1[i];
	}
	ll ans = 0, tempmax = 0;
	for (int i = 0; i < n-1; i++)
	{
		if (n1[i] >= n1[i + 1])
			tempmax++;
		else
		{
			ans = max(tempmax, ans);
			tempmax = 0;
		}
	}
	ans = max(tempmax, ans);
	cout << ans << endl;
	return 0;
}*/

int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int a, b;
	while (cin >> a >> b)
	{
		int sum = 0;
		if (b == 1)
		{
			cout << 0 << endl;
			continue;
		}
		else
		{
			if (b <= a)
			{
				cout << 1 << endl;
				continue;
			}
			else
			{
				b -= a;
				sum++;
				while (b > 0)
				{
					sum++;
					b -= a - 1;
				}
				cout << sum << endl;
			}
		}
		
	
	}
	return 0;
}