tape ——cf

B. Tape

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You have a long stick, consisting of mm segments enumerated from 11 to mm . Each segment is 11 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 tt placed at some position ss will cover segments s,s+1,…,s+t−1s,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 kk 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 nn , mm and kk (1≤n≤1051≤n≤105 , n≤m≤109n≤m≤109 , 1≤k≤n1≤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,…,bnb1,b2,…,bn (1≤bi≤m1≤bi≤m ) — the positions of the broken segments. These integers are given in increasing order, that is, b1<b2<…<bnb1<b2<…<bn .

Output

Print the minimum total length of the pieces.

Examples

Input

Copy

4 100 2
20 30 75 80

Output

Copy

17

Input

Copy

5 100 3
1 2 4 60 87

Output

Copy

6

Note

In the first example, you can use a piece of length 1111 to cover the broken segments 2020 and 3030 , and another piece of length 66 to cover 7575 and 8080 , for a total length of 1717 .

In the second example, you can use a piece of length 44 to cover broken segments 11 , 22 and 44 , and two pieces of length 11 to cover broken segments 6060 and 8787 .

就是有n个洞，消费为n，要消减成k，所有把n-k个两个洞合并，合并之后就会变成k堆

 

 

就是你先贴上，花费为n，你要把花费减到k，就是要合并n-k个两个洞的距离，差分出来取最小的n-k个再加上单独补
两个洞的pos差分是等于补一个洞和中间段的

，这时候就补上这k个洞即可。

#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+100;
int a[maxn],b[maxn];
int main()
{
    int n,m,k;
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=1;i<=n-1;i++) b[i]=a[i+1]-a[i];
    sort(b+1,b+n);
    ll sum=0;
    for(int i=1;i<=n-k;i++) sum+=b[i];
    sum+=k;
    printf("%I64d\n",sum);
    return 0;
}