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)

//#pragma GCC target ("avx2")
//#pragma GCC optimization ("O3")
//#pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>
using namespace std;

#define mem(arr, i) fill(arr, arr + sizeof(arr), i)
#define MEM(arr, i) memset(arr, i, sizeof(arr))
#define Push push_back
#define Pair make_pair
#define ALL(x) x.begin(), x.end()
#define X first
#define Y second
#define ACCEL \
	ios_base::sync_with_stdio(false); \
	cin.tie(NULL); \
	cout.tie(NULL)
typedef unsigned long long ull;
typedef long long lli;
typedef pair<int,int> pii;
typedef pair<lli,lli> pll;
typedef pair<pll,lli> p3l;
typedef vector<int> vcI;
typedef vector<lli> vcL;
typedef vector<pii> vPI;
typedef vector<pll> vPL;
typedef vector<p3l> v3L;
typedef vector<vcL> vvL;
const double PI = 3.141592653589793;
const lli I_MAX = 1LL << 60;
      lli   mod = 1000000007;

/// templates ///

const lli maxN = 2E5;

vcL inv(maxN+5, 1);

void ADD(lli &x, lli y) {
	x += y;
	if (x >= mod) x -= mod;
	if (x < 0) x += mod;
}

void MUL(lli &x, lli y) {
	x *= y;
	x = (x % mod + mod) % mod;
}

lli mypow(lli b, lli e) {
	if (b == 0) return 0;
	lli ans = 1;
	while (e) {
		if (e & 1) MUL(ans, b);
		MUL(b, b);
		e >>= 1;
	}
	return ans;
}

lli modinv(lli n) {
	return mypow(n, mod - 2);
}

void calcInv(lli n) {
	for (int i = 2; i <= n; ++i) {
		inv[i] = (mod - mod / i) * inv[mod % i] % mod;
	}
}

/// coding area ///

const lli maxn = 1E5;

void solve() {
	lli a, b;
	cin >> a >> b;
	cout << (b - 2) / (a - 1) + 1 << '\n';
}

int main() {
//	precal();
	int t = 1;
	#ifdef LOCAL
		t = INT_MAX;
	#else
		ACCEL;
	#endif
//	cin >> t;
	for (int i = 1; i <= t; ++i) {
//		cout << "Case #" << i << ": ";
		solve();
	}
	return 0;
}