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;
typedef long long ll;
#define rep(i, n) for(ll i = 0; i < (int)(n); i++)

ll max(ll a, ll b){ 
	if(a >= b) return a;
	else if(a < b) return b;
}
ll min(ll a, ll b){ 
	if(a <= b) return a;
	else if(a > b) return b;
}
ll gcd(ll a, ll b){
	if(b == 0) return a;
	return gcd(b, a%b);
}
ll fact(ll n){
	ll res = 1;
	rep(i, n){
		res *= i + 1;
		res =  res%((ll)(1e9 + 7));
	}
	return res;
}
ll digit(const ll n){
	ll res = 0;
	ll i = 1;
	while(n % i != n){
		res += 1;
		i *= 10;
	}
	return res;
}

int main(){
	ll a, b;
	cin >> a >> b;
	ll ans = 1;
	ans += (b-a)/(a-1);
	cout << ans << endl;
}