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: Wrong Answer)

#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
#include<math.h>
#include<string>
#include <iomanip>
#include<deque>
#include<queue>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef int itn;
const ll LINF = 1e18;
const int INF = 1e8;

//マクロ定義
#define vvint(vec,n,m,l) vector<vector<int>> vec(n, vector<int>(m,l));	// lで初期化
#define vvll(vec,n,m,l) vector<vector<ll>> vec(n,vector<ll>(m,l));
#define vint vector<int>
#define pint pair<int,int>
#define rep(i,a) for(int i=0;i<(a);i++)
#define all(x) (x).begin(),(x).end()
#define debug system("pause")				//デバッグ用


int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int a, b;
	cin >> a >> b;

	int kuti = a;
	int ko = 1;
	while (true)
	{
		if (kuti >= b)
		{
			break;
		}
		kuti--;
		kuti += a;
		ko++;
	}

	cout << ko << endl;
}