B. Tape(贪心）

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 7575and 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.

题意：在一个长度为m的区间内有n处破损，我们要至多k次使用胶带将其全部填补。

正向考虑：

#include<bits/stdc++.h>
using namespace std;
int n,k,m,x,y,a[100010],ans;
int main()
{
    cin>>n>>m>>k;
    for(int i=1;i<=n;i++)
    {
        cin>>x;
        a[i]=x-y-1;//将间隔分别填到数组里
        y=x;
    }
    sort(a+2,a+1+n);    //从第二个间隔起让其排序，因为第一个间隔无论多大，都不影响最优值
    for(int i=2;i<=n-k+1;i++)//一开始我们有n段，但至多用k段，我们要想办法融合其中间隔将其变成k段（n->k为（n-k））即n-（n-k）=k
        ans+=a[i];
    cout<<n+ans<<endl;  //n是每个点都得用1个Tape，ans是其余最佳的Tape
    return 0;
}

反向考虑：

#include<bits/stdc++.h>
using namespace std;
int n,k,m,a[100010],ans,cnt;
//priority_queue<int,vector<int>,greater<int> >que;//从小到大出堆
priority_queue<int,vector<int>,less<int> >que;//从大到小出队，由于默认为最大堆，这与priority_queue<int>que等价

int main()
{
    cin>>n>>m>>k;
    for(int i=1;i<=n;i++)
        cin>>a[i];
    for(int i=2;i<=n;i++)
    {
        int d=a[i]-a[i-1]-1;
        que.push(d);
    }
    cnt=1;          //一开始将从第一个破损处到第n个破损处全部覆盖，为1段
    ans=a[n]-a[1]+1;
    while(cnt<k)
    {
        ans-=que.top();  //ans减去最大的即为最小的
        que.pop();
        cnt++;
    }
    cout<<ans<<endl;
    return 0;
}