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 <bits/stdc++.h>
#include <math.h>
#include <algorithm>
#define ll long long
#define ld long double
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pld pair<ld, ld>
#define f first
#define s second
#define fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define pb push_back
#define eb emplace_back
#define in insert
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define FOR(i, n, x) for(int i = x; i < n; i++)
#define OUT(a, e) FOR(i, e, 0) cout << a[i] << ' '
#define OUTV(a) for(auto i : a) cout << i << ' ';
#define maxim(a, n, mx) FOR(i, n, x)mx = max(mx, a[i]);



using namespace std;



int main()
{
    int a, b; cin >> a >> b;
    int now = 1, cnt = 0;
    while(now < b)
    {
        now--; now += a; cnt++;
    }
    cout << cnt;
    return 0;
}