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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder_solve
{
    class Program
    {
        public const ulong mod = 1000000007;

        static void Main(string[] args)

        {
            var ab = Input3();
            var ans = 1;
            var sct = ab[0];

            while (sct < ab[1])
            {
                sct += ab[0] - 1;
                ans++;
            }
            

            Console.WriteLine(ans);
                
                





        }

       
        static public long Input()
        {
            return long.Parse(Console.ReadLine());
        }

        static public string[] Input2()
        {
            return Console.ReadLine().Split();
        }

        static public long[] Input3()
        {
            string[] data = Console.ReadLine().Split();
            long[] res = new long[data.LongLength];
            for(long i = 0; i < data.LongLength; i++)
            {
                res[i] = long.Parse(data[i]);
            }
            return res;
        }
        public static long Choose(long n, long a)
        {
            long x = 1;
            long y = 1;
            for (long i = 0; i < a; i++)
            {
                x *= n - i;
                y *= i + 1;
            }
            return x / y;
        }

        public static long Gcd(long a, long b)
        {
            long x = Math.Max(a, b);
            long y = Math.Min(a, b);
            if (x % y == 0) return y;
            else return Gcd(y, x % y);
        }

        
       
    }
}