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<bits/stdc++.h>
#include<algorithm>
#include<math.h>
#include<iostream>
using namespace std;
long long int lcm(long long int a,long long int b){
  long long int n;
  n=a*b/__gcd(a,b);
  return n;
}
using ll =long long int;
#define ci(a) cin>>a
#define co(a) cout<<a<<endl
#define gcd(a,b) __gcd(a,b)
#define vec vector<ll>
#define for(i,a,b) for(int i=a;i<b;i++)
#define sort(a,b) sort(a.begin(),b.end())
//gcd 最大公約数
//lcm 最小公倍数
//double 少数（変数）
//string 文字列（変数）
//sqrt 平方根
//ceil 天井関数
//floor 床関数
//abs 絶対値
//pow 累乗
//cbrt  立方根
//hypot 平方和の平方根
//(string-'0') 文字列を数字に変換
//vector<vector<int>> data(a,vector<b>(b)) a×b行列
//for(int i=0;i<a;i++)
//for(int j=0;j<b;j++)
//cin>>data.at(i).at(j);a×b行列の入力
 
int main() {
  ll a,b;
  ci(a>>b);
  ll c=1;
  for(i,1,50){
    c+=a-1;
    if(c>=b){
      co(i);
      break;
    }
  }
}