Codeforces Global Round 1 B. Tape (贪心)

                                                                                  B. Tape

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 tt placed at some position ss 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 nn , mm and kk (1≤n≤105, n≤m≤109n≤m≤109 , 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

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 11 to cover the broken segments 20 and 30 , and another piece of length 66 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 , 2and 4 , and two pieces of length 1 to cover broken segments 60 and 87 .

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define Swap(a,b)  a ^= b ^= a ^= b
#define pi acos(-1)
#define cl(a,b) memset(a,b,sizeof(a))
using namespace std ;
typedef long long LL;
const int N = 1e7+10 ;
const int inf = 0x3f3f3f3f;
const int MAX = 1e5+5;
int a[MAX];
int d[MAX];
bool cmp(const int &a , const int &b){
	return a>b ;
}
int main()
{
	int n , m , k ;
	int ans ;
	scanf("%d%d%d",&n,&m,&k);
	for(int i = 1 ; i <=n ; i++ ){
		cin >>a[i] ;
	}
	for(int i = 1 ; i