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)

#pragma GCC optimize (3)
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define PII pair<int,int>
#define ll long long
#define ull unsigned long long
#define INFI 2147483647
#define INFL 9223372036854775807
#define INFU 18446744073709551615
using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-6;
inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	while (ch<'0' || ch>'9') {
		if (ch == '-') f = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return x * f;
}
char s[3], t[3];
int main()
{
	//ios_base::sync_with_stdio(false);
	//cin.tie(0);
	//cout.tie(0);
	int a, b;
	while (scanf("%d%d", &a, &b) == 2) {
		if (a >= b) cout << 1 << endl;
		else {
			int res = 1;
			while (a*res < b) res++;
			cout << res << endl;
		}
	}
	return 0;
}