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<cstdlib>
#include <cmath>

#define INF 100000000
template<class T> inline bool chmax(T& a, T b)
{ if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b)
{ if (a > b) { a = b; return 1; } return 0; }

using namespace std;

int main(){
  int i,j,k,n,d,a[100],x,y,z=1,ans=0;

  cin >>x>>y;
  for(i=0;i<21;i++){

    if(i==1)a[i]=x;
    else{
      a[i]=(x*i)-(i-1);
    }
  }

  for(i=0;i<21;i++){
    if(y<=a[i])break;
  }

  cout << i <<endl;

  return 0;
}