You wish to buy video games from the famous online video game store Mist.

Usually, all games are sold at the same price, $\boldsymbol{p}$ dollars. However, they are planning to have the seasonal Halloween Sale next month in which you can buy games at a cheaper price. Specifically, the first game will cost $\boldsymbol{p}$ dollars, and every subsequent game will cost $\boldsymbol{d}$ dollars less than the previous one. This continues until the cost becomes less than or equal to $m$ dollars, after which every game will cost $m$ dollars. How many games can you buy during the Halloween Sale?

Example

$p=20$

$\boldsymbol{d}=3$

$m=6$

$s=70$.

The following are the costs of the first $\mbox{11}$, in order:

$20,17,14,11,8,6,6,6,6,6,6$

Start at $p=20$ units cost, reduce that by $\boldsymbol{d}=3$ units each iteration until reaching a minimum possible price, $m=6$. Starting with $s=70$ units of currency in your Mist wallet, you can buy 5 games: $20+17+14+11+8=70$.

Function Description

Complete the howManyGames function in the editor below.

howManyGames has the following parameters:

int p: the price of the first game
int d: the discount from the previous game price
int m: the minimum cost of a game
int s: the starting budget

Input Format

The first and only line of input contains four space-separated integers $\boldsymbol{p}$, $\boldsymbol{d}$, $m$ and $\boldsymbol{\mathrm{~s~}}$.

Constraints

$1\leq m\leq p\leq100$
$1\leq d\leq100$
$1\leq s\leq10^4$

Sample Input 0
20 3 6 80

Sample Output 0
6

Explanation 0

Assumptions other than starting funds, $\boldsymbol{\mathrm{~S~}}$, match the example in the problem statement. With a budget of $\textbf{80}$, you can buy $\boldsymbol{6}$ games at a cost of $20+17+14+11+8+6=76$. A $7^{th}$ game for an additional $\boldsymbol{6}$ units exceeds the budget.

Sample Input 1
20 3 6 85

Sample Output 1
7

Explanation 1

This is the same as the previous case, except this time the starting budget $s=85$ units of currency. This time, you can buy $7$ games since they cost $20+17+14+11+8+6+6=82$. An additional game at $\boldsymbol{6}$ units will exceed the budget.

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