#define _CRT_SECURE_NO_WARNINGS
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC optimize ("O3")
//#pragma GCC target("avx2")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")
//#pragma GCC target("bmi,bmi2,popcnt,lzcnt")
//Babahnineeleven will win IOI 2040
//Tanya Zadniprovska will win EGOI 2025
//Andrew Holod will NOT win IOI 2025
//Andrew Holod did not qualify to IOI 2025))))))))))
//Andrew Pavlyk is best coder in Khmelnytskiiii
#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cmath>
#include <fstream>
#include <climits>
#include <queue>
#include <algorithm>
#include <stdint.h>
#include <stack>
#include <iomanip>
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <cstring> // �л� memset
using namespace std;
#define endl '\n'
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef std::pair <ll, ll> pii;
typedef std::pair <ll, ull> piu;
typedef std::pair <ld, ld> pdd;
const ll DIM = 300007;
const ld PI = 3.14159265358979;
const ll SQDIM = 460;
const ll MXMASK = (1 << 20);
const ll INF = 1e18;
const ll mod = 1e9 + 7;
const ld eps = 0.00000000001;
const ll Bits = 32;
const ll Bsize = 300;
const int dx[4] = { 1, 0, -1, 0 };
const int dy[4] = { 0, 1, 0, -1 };
FILE* stream;
ll n, k;
ll a[DIM];
bool operator<(pii p1, pii p2) {
if (p1.first != p2.first) return p1.first < p2.first;
return p1.second > p2.second;
}
pii solve(ll p) {
vector < pii > F(n + 1);
pii mxF = { 0, 0 };
pii best = { 0, 0 };
for (int i = 1; i <= n; i++) {
F[i] = max(F[i - 1], { best.first + a[i] - p, best.second + 1 });
mxF = max(F[i], mxF);
best = max(best, { mxF.first - a[i], mxF.second });
}
return F[n];
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifdef _DEBUG
freopen_s(&stream, "input.txt", "r", stdin);
freopen_s(&stream, "output.txt", "w", stdout);
#endif
cin >> n >> k;
for (int i = 1; i <= n; i++) cin >> a[i];
for (int i = 1; i <= n; i++) a[i] += a[i - 1];
ll L = 0, R = INF;
while (R - L > 1) {
ll mid = (L + R) >> 1;
if (solve(mid).second <= k) R = mid;
else L = mid;
}
cout << solve(R).first + solve(R).second * R << endl;
return 0;
}
