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<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<map>
#include<queue>
#include<deque>
#include<iomanip>
#include<tuple>
using namespace std;
typedef long long int LL;
typedef pair<int,int> P;
typedef pair<LL,int> LP;
const int INF=1<<30;
const LL MAX=1e9+7;

void array_show(int *array,int array_n,char middle=' '){
	for(int i=0;i<array_n;i++)printf("%d%c",array[i],(i!=array_n-1?middle:'\n'));
}
void array_show(LL *array,int array_n,char middle=' '){
	for(int i=0;i<array_n;i++)printf("%lld%c",array[i],(i!=array_n-1?middle:'\n'));
}

int main(){
	int n,m;
	int i,j,k;
	int a,b,c;
	cin>>a>>b;
	a--,b--;
	a=(b-1)/a+1;
	cout<<a<<endl;
}