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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define ll long long
#define rep(i,l,r) for(ll i=(l);i<(r);i++)
#define max(p,q) ((p)>(q)?(p):(q))
#define min(p,q) ((p)<(q)?(p):(q))
#define abs(p) ((p)>=(0)?(p):(-(p)))
int cmp(const void *a, const void *b);
int main(void){
    ll A, B, tap;
    scanf("%lld %lld", &A, &B);
    if(B==0){
        printf("0\n");
        return 0;
    }
    for(tap=1; 1+A*tap-tap<B; tap++);
    printf("%lld\n", tap--);
    return 0;
}
int cmp(const void *a, const void *b){
    ll A=*(ll *)a, B=*(ll *)b;
    if(A==B)return 0;
    else return A>B ? 1:-1;//昇順ソート
}