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)

/*Remember: The Struggle Is Important*/

//CF,CC,AtC,SPOJ: hp1999
//HE: hemant269
//HR: hemant2132

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

#define     int          long long int
#define     fast()       ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define     all(x)       x.begin(),x.end()
#define     mem(a,b)     memset(a,b,sizeof(a))
#define     gcd(a,b)     __gcd((a),(b))
#define     lcm(a,b)     ((a)*(b))/gcd((a),(b))
#define     pb           push_back
#define     ins          insert
#define     F            first
#define     S            second

const int inf=1e18,M=1e9+7;
const int N=1;

void solve()
{
    int a,b;
    cin>>a>>b;

    if(b==1)
    {
        cout<<0;
        return;
    }

    --a;
    int ans=b/a;

    if(!ans)
    {
        cout<<1;
        return;
    }

    if(b%a>1)
        ++ans;
    cout<<ans;
}

int32_t main()
{
    fast();

    int t=1;
    //cin>>t;
    while(t--)
    {
        solve();
    }

    return 0;
}