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"
using namespace std;
#define INF 1000000000
#define rep(i, n) for(int i = 0; i < n; i++)
#define repr(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define ALL(x) (x).begin(), (x).end()  //昇順
#define RALL(x) (x).rbegin(), (x).rend() // 降順
#define SZ(x) ((int)(x).size)
const long long mod = 1e9+7;
typedef priority_queue<int> PQ;
typedef vector<long long> Vl;
typedef vector<int> VI;  // VI a(n);
typedef vector<string> VS;  
typedef vector<char> VC; 
typedef vector<VS> VSS;
typedef vector<VC> VCC;    
typedef vector<VI> VVI;   // VVI a(n,vector<int>(m)) n * m 
typedef vector<Vl> VVl;
typedef pair<int, int> PII;
typedef map<int, int> MP;
typedef vector<pair<int, int> > PS;
typedef long long ll;  //ll とdoubleは違う
// cout << << endl;
// cin >> ;
template<typename T>   // operator << (cout,a);
ostream& operator << (ostream& os, vector<T>& v) {
    os << "{";
    rep(i, (int)v.size()) { os << v[i] << (i < v.size() - 1 ? ", " : ""); }
    os << "}";
    return os;
}

int main(){
    ll a,b,x=0,y,d=0,n,h,w,k,m,min=1000,ans=0,ptr=0,sum=0;
    string s,s1,s2;
    bool f=true;
    cin >> a >> b;
    while(f){
        if(sum+a>=b){
            ans++;
            f = false;
        }
        else if(b > sum){
            sum += a-1;
            ans++;
        }
    }
    cout << ans << endl;
    return 0;
}