KEYENCE is famous for quick delivery.

In this problem, the calendar proceeds as Day $1$, Day $2$, Day $3$, $\dots$.

There are orders $1,2,\dots,N$, and it is known that order $i$ will be placed on Day $T_i$.
For these orders, shipping is carried out according to the following rules.

• At most $K$ orders can be shipped together.
• Order $i$ can only be shipped on Day $T_i$ or later.
• Once a shipment is made, the next shipment cannot be made until $X$ days later.
  • That is, if a shipment is made on Day $a$, the next shipment can be made on Day $a+X$.

For each day that passes from order placement to shipping, dissatisfaction accumulates by $1$ per day.
That is, if order $i$ is shipped on Day $S_i$, the dissatisfaction accumulated for that order is $(S_i - T_i)$.

Find the minimum possible total dissatisfaction accumulated over all orders when you optimally schedule the shipping dates.

The input is given from Standard Input in the following format:

$N$ $K$ $X$

$T_1$ $T_2$ $\dots$ $T_N$

• All input values are integers.
• $1 \le K \le N \le 100$
• $1 \le X \le 10^9$
• $1 \le T_1 \le T_2 \le \dots \le T_N \le 10^{12}$

For example, by scheduling shipments as follows, we can achieve a total dissatisfaction of $2$, which is the minimum possible.

• Ship order $1$ on Day $1$.
  • This results in dissatisfaction of $(1-1) = 0$, and the next shipment can be made on Day $4$.
• Ship orders $2$ and $3$ on Day $6$.
  • This results in dissatisfaction of $(6-5) + (6-6) = 1$, and the next shipment can be made on Day $9$.
• Ship order $4$ on Day $10$.
  • This results in dissatisfaction of $(10-10) = 0$, and the next shipment can be made on Day $13$.
• Ship order $5$ on Day $13$.
  • This results in dissatisfaction of $(13-12) = 1$, and the next shipment can be made on Day $16$.