You have a long stick, consisting of mm 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,…,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?
Input
The first line contains three integers n, m and k (1≤n≤1e5, n≤m≤1e9, 1≤k≤n) — the number of broken segments, the length of the stick and the maximum number of pieces you can use.
The second line contains nn integers b1,b2,…,bn (1≤bi≤m) — the positions of the broken segments. These integers are given in increasing order, that is, b1<b2<…<bn.
Output
Print the minimum total length of the pieces.
Examples
Input
4 100 2
20 30 75 80
Output
17
Input
5 100 3
1 2 4 60 87
Output
6
Note
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 11 to cover broken segments 60 and 87.
题意：n的断点，k次机会（必须用完）来粘贴上断点，求粘贴的最小长度
题解：粘贴上距离为 s 的两个断点，用的长度是 s+1；如果有三个断点，比如 1，2，4；可以用长度为4的胶布来粘贴1-4，也可以用长度为2粘贴2-4、长度为3来粘贴2-4，总的为5；
--->不管一段距离内你想粘贴p次胶布，和你一次粘贴相隔的就是（p-1）长的距离
所以说：用的胶布距离应该这样计算比较方便：粘贴s到d，胶布为d-s+k（k是你这个过程中 分开几次 粘贴）
所以思路为：把每2个 相邻 断点之间的距离求出来建成新的数组bn，排序后求出最小的几个距离即可，最后的ans加上k
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
const int max_n=1e5+5;
LL n,m,k,ans,a[max_n],b[max_n];
int main()
{
//freopen("1.in","r",stdin);
//freopen("1.out","w",stdout);
scanf("%lld%lld%lld",&n,&m,&k);
for(LL i=0;i<n;i++)
{
scanf("%lld",&a[i]);
}
for(LL j=0;j<n-1;j++)
{
b[j]=a[j+1]-a[j];
}
sort(b,b+n-1);
for(LL i=0;i<n-k;i++)
{
ans+=b[i];
}
printf("%lld",ans+k);
return 0;
}