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>
using namespace std;
#define ll long long
#define pll pair<long long,long long>
#define mod 1000000007
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL);

ll power(ll a,ll b){
	ll ans=1;
	while(b>0){
		if(b%2!=0){
			ans=(ans*a)%mod;
		}
		a=(a*a)%mod;
		b>>=1;
	}
	return ans;
}

int main() {
	IOS;
	int t=1;
	//cin>>t;
	while(t--){
		ll p,q;
      cin>>p>>q;
      cout<<ceil(q/double(p));
	}
	return 0;
}