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 IOS ios::sync_with_stdio(false); cin.tie(0);
#define FOR(i,s,n) for(int i = (s); i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) for(int i = (n); i >= 0; i--)
#define ALL(n) (n).begin(), (n).end()
#define RALL(n) (n).rbegin(), (n).rend()
#define ATYN(n) cout << ( (n) ? "Yes":"No") << '\n';
#define CFYN(n) cout << ( (n) ? "YES":"NO") << '\n';
#define OUT(n) cout << (n) << '\n';
using ll = long long;
using ull = unsigned long long;
using pii = pair<int,int>;
using pll = pair<ll,ll>;

int main(void)
{
    IOS
    int a, b;
    cin >> a >> b;
    int ans = 1;
    int now = a;
    while(now < b) {
        now += a-1;
        ans++;
    } 

    cout << ans << '\n';

    return 0;
}