This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <iomanip>
#include <cassert>
#include <stack>
#include <queue>
#include <deque>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
//using namespace __gnu_pbds;
typedef long long ll;
typedef pair <int, int> pii;
// template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// order_of_key (k) : Number of items strictly smaller than k .
// find_by_order(k) : K-th element in a set (counting from zero).
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define pb push_back
#define ppb pop_back
#define mkp make_pair
#define F first
#define S second
#define show(a) cerr << #a <<" -> "<< a <<"\n"
#define fo(a, b, c, d) for(int (a) = (b); (a) <= (c); (a) += (d))
#define foo(a, b, c ,d) for(int (a) = (b); (a) >= (c); (a) -= (d))
//#define int ll
const int N = 2e5 + 5;
const int INF = 1e9 + 5;
int n, k, ans, t[N];
vector <int> vec;
void solve() {
cin >> n >> k;
for (int i = 1; i <= n; ++i) {
cin >> t[i];
if (1 <= i-1) vec.pb(t[i] - (t[i-1]+1));
}
int ans = t[n] + 1 - t[1];
sort(all(vec));
reverse(all(vec));
for (int i = 0; i < k-1; ++i) {
ans -= vec[i];
}
cout << ans;
}
int main () {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int tt = 1;
while (tt --) {
solve();
}
return 0;
}
/*
If you only do what you can do,
You will never be more than you are now!
*/
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