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 <iomanip>
#include <bits/stdc++.h>
#include <vector>
#include <iterator>
#include <string>
#define pan(i,n) for(int i=0;i<n;i++)
#define ll long long int
#define vi vector<int>
#define pb push_back
using namespace std;



int main(){
    int i,a,b;
    cin>>a>>b;
    int sok=0;
    if(b!=1){
        for( i=1;i<29;i++){
            sok=(a*i)+1-i;
            // cout<<sok;
            if(sok>=b) break;
        }
    }
    if(sok>1) cout<<i;
    else cout<<0;
}