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)

//ehhhhh...... i hate you......

/* optimizations and other useless stuff*/
#pragma warning(disable:4996)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize ("-ffloat-store")
#include<bits/stdc++.h>
#define FastIO ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define MOD 1000000007
#define INF 1e14
#define ull unsigned long long int
using namespace std;
/* TYPE DEFINITIONS */
typedef     long long int   ll;
typedef     vector<ll>      vi;
typedef     pair<ll,ll>     pi;


/* Basic  Math!!! */
bool isPrime(ull n) 
{ 
    if (n <= 1)  return false; 
    if (n <= 3)  return true; 
    if (n%2 == 0 || n%3 == 0) return false; 
  
    for (ull i=5; i*i<=n; i=i+6) 
        if (n%i == 0 || n%(i+2) == 0) 
           return false; 
  
    return true; 
} 
ll ncr(int n, int k)  
{  
    ll res = 1;  
    if ( k > n - k )  
        k = n - k;  
    for (int i = 0; i < k; ++i)  
    {  
        res *= (n - i);  
        res /= (i + 1);  
    }  
  
    return res;  
} 

/* MACROS */
#define  all(A)         (A).begin() , (A).end()
#define  rall(A)        (A).rbegin() , (A).rend()
#define  sz(A)          (int)(A).size()
#define  pb             push_back
#define  ppb            pop_back
#define  mp             make_pair
#define  ln(X)          (int)(X).length()
#define  square(X)      ((X)*(X))
#define  cube(X)        ((X)*(X)*(X))
#define  forn(i, n)  for (int i = 0; i < int(n); i++)
#define  forr(i, n)  for (int i = int(n - 1); i >= 0; i--)
#define  fora(i, a, b)  for (int i = int(a); i <= int(b); i++)
#define  forb(i, a, b)  for (int i = int(a); i >= int(b); i--)
#define  fore(it, a)    for(__typeof((a).sbegin()) it = (a).begin(); it != (a).end(); it++)  
void solve()
{
    int a,b,s=0,c=0;
    cin>>a>>b;
    while(s<=b)
    {
        s+=a;
        c++;
        if(s>=b)break;
        s--;
    }
    cout<<c<<endl;
}

int main()
{
    FastIO;
    ll c=0;
    ll t=1;
    //cin>>t;
    while(t--)
    {
        //cout<<"Case "<<++c<<": ";
        solve();
    }
    return 0;
}