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)

/** 
 * Dont raise your voice, improve your argument.
 * --Desmond Tutu
 */

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define endl "\n"
#define fori(n) for(ll i=0; i<(n); i++)

#define cins(s) string s; cin>>s;
#define cini(i) int i; cin>>i;
#define cinll(l) ll l; cin>>l;
#define cinai(a, n) vi a(n); fori(n) { cin>>a[i]; }
#define cinall(a, n) vll a(n); fori(n) { cin>>a[i]; }
#define cinas(s, n) vs s(n); fori(n) { cin>>s[i]; }

typedef pair<int, int> pii;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<string> vs;
typedef vector<ll> vll;

int main() {
cin.tie(nullptr);
std::ios::sync_with_stdio(false);
    cini(a);
    cini(b);

    int ans=0;
    int s=1;
    while(s<b) {
        s+=(a-1);
        ans++;
    }

    cout<<ans<<endl;

}