CF1110B.Tape

You have a long stick, consisting of $m$ segments enumerated from $1$ to $m$. Each segment is $1$ centimeter long. Sadly, some segments are broken and need to be repaired.

You have an infinitely long repair tape. You want to cut some pieces from the tape and use them to cover all of the broken segments. To be precise, a piece of tape of integer length $t$ placed at some position $s$ will cover segments $s, s+1, \ldots, s+t-1$.

You are allowed to cover non-broken segments; it is also possible that some pieces of tape will overlap.

Time is money, so you want to cut at most $k$ continuous pieces of tape to cover all the broken segments. What is the minimum total length of these pieces?

你有一根长棍，由 $m$ 个编号从 $1$ 到 $m$ 的小段组成。每段长度为 $1$ 厘米。不幸的是，其中一些小段已损坏，需要修复。

你拥有一卷无限长的修复胶带。你想从胶带上裁剪出若干段，并用它们覆盖所有损坏的小段。具体而言，一段长度为整数 $t$ 的胶带若起始位置为 $s$，则它将覆盖小段 $s, s+1, \ldots, s+t-1$。

允许胶带覆盖未损坏的小段；也允许不同胶带段之间相互重叠。

时间就是金钱，因此你最多只能裁剪 $k$ 段连续的胶带以覆盖所有损坏的小段。这些胶带段的总长度最小是多少？

The first line contains three integers $n$, $m$ and $k$ ($1 \le n \le 10^5$, $n \le m \le 10^9$, $1 \le k \le n$) — the number of broken segments, the length of the stick and the maximum number of pieces you can use.

The second line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le m$) — the positions of the broken segments. These integers are given in increasing order, that is, $b_1 \lt b_2 \lt \ldots \lt b_n$.

第一行包含三个整数 $n$、$m$ 和 $k$（$1 \le n \le 10^5$，$n \le m \le 10^9$，$1 \le k \le n$）—— 分别表示断裂点的数量、木棍的长度以及最多可使用的段数。

第二行包含 $n$ 个整数 $b_1, b_2, \ldots, b_n$（$1 \le b_i \le m$）—— 表示各断裂点的位置。这些整数按严格递增顺序给出，即 $b_1 \lt b_2 \lt \ldots \lt b_n$。

Print the minimum total length of the pieces.

输出各段的最小总长度。

In the first example, you can use a piece of length $11$ to cover the broken segments $20$ and $30$, and another piece of length $6$ to cover $75$ and $80$, for a total length of $17$.

In the second example, you can use a piece of length $4$ to cover broken segments $1$, $2$ and $4$, and two pieces of length $1$ to cover broken segments $60$ and $87$.

在第一个例子中，你可以使用一段长度为 $11$ 的修补材料覆盖破损段 $20$ 和 $30$，再用一段长度为 $6$ 的修补材料覆盖破损段 $75$ 和 $80$，总长度为 $17$。

在第二个例子中，你可以使用一段长度为 $4$ 的修补材料覆盖破损段 $1$、$2$ 和 $4$，再用两段长度均为 $1$ 的修补材料分别覆盖破损段 $60$ 和 $87$。