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>

#define rep(i,n) for(int i=0; i<(n); i++)
#define reps(i,n) for(int i=1; i<=(n); i++)
#define pii pair<int,int>
#define gsort(x,n) sort((x),(x)+(n),greater<int>())
#define all(x) (x).begin(), (x).end()
#define int long long
#define abs(a) llabs(a)
#define re(s) reverse((s).begin(),(s).end())
#define setout(n,x) setw(n) << setfill(x)
#define Fixed fixed << setprecision(10)
const long long mod = 1000000007;

using namespace std;
long long h[1000010],x[1010][1010],z[1000010];
int a,b,c,n,cnt=0,ans=0;
signed main(){

    cin.tie(0);
    ios::sync_with_stdio(false);
    
    int a,b;
    cin>>a>>b;
    int cnt=0,ans=0;

    while(1){
        ans++;
        cnt=cnt+a;
        if(cnt>=b) break;
        cnt--;
    }
    cout<<ans<<'\n';
}