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 <algorithm>
#include <string>
#include <map>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <fstream>
#include <queue>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <sstream>
#include <math.h>
#include <functional>

#define For(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  For(i,0,n)

//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))
#define SORT(c) sort((c).begin(),(c).end())
#define vi(m,a) vector<int> m(a)
#define vti(m,a,i) vector<vector<int>> m(a,vector<int>(i))
#define ALL(it,a) for(auto it = a.begin(); it!=a.end(); it++)
#define Fe(it,a) for(auto &it : a)
#define all(a) begin(a),end(a)

using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef pair<ll, ll> pll;
#define WARU 1000000007

int dx[] = { 1,0,-1,0 };
int dy[] = { 0,1,0,-1 };

int main() {
	int a, b;
	cin >> a >> b;
	if (b == 1) {
		cout << 0 << endl;
		return 0 ;
	}
	int ko = 1;
	int result = 0;
	rep(i, 100) {
		if (b <= (a - 1) * (i - 1) + a) {
			result = i;
			break;
		}
	}
	cout << result << endl;


}