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)

/*
　　こんにちは。あたしはｶｳｶﾞｰﾙ。
　　redcoderになるためAtCoderを巡る旅をしてます。

　　　 　 ＿_
　　　 ヽ|＿_|ノ　　　　ﾓｫ
　　　　||‘‐‘||ﾚ　　_)_, ―‐ 、
　　　　/(Ｙ (ヽ＿ /・ ヽ　　 ￣ヽ
　　　　∠ ＿ ゝ　 ｀^ヽ ﾉ.::::_(ノヽ
　　　　 _/ヽ　 　　  /ヽ￣￣/ヽ
*/

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define dump(x) cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE") << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << endl

#define rep(i, n) REP(i, 0, n)                  // 0, 1, ..., n-1
#define REP(i, x, n) for(int i = x; i < n; i++) // x, x + 1, ..., n-1
#define FOREACH(x, a) for(auto &(x) : (a))

#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()

#define COUT(x) cout << (x) << endl

template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return 1;
    }
    return 0;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int a, b, ans;
    cin >> a >> b;
    for(int i = 1;; i++) {
        if((a * i - (i - 1)) >= b) {
            ans = i;
            break;
        }
    }
    if(b == 1) { ans = 0; }
    COUT(ans);
}