#include<bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using ll = long long;
using vll = vector<ll>;
using ii = pair<int, int>;
using vii = vector<ii>;
using ld = long double;
#define rep(i, a, b) for (int i = a; i < b; i++)
#define all(x) x.begin(), x.end()
#define pb(x) push_back(x)
#define sz(x) x.size()
pair<ld, int> f(vector<ld> &nums, ld x) {
int n = nums.size();
vector<ld> pf(n), dp(n);
vi used(n);
int best = 0;
rep(i, 1, n) {
pf[i] = pf[i - 1] + nums[i];
dp[i] = pf[i] + dp[best] - pf[best] - x;
used[i] = used[best] + 1;
if (dp[i - 1] > dp[i]) {
dp[i] = dp[i - 1];
used[i] = used[i - 1];
}
if (dp[i] - pf[i] > dp[best] - pf[best])
best = i;
}
best = 0;
rep(i, 1, n)
if (dp[i] > dp[best])
best = i;
return {dp[best], used[best]};
}
int main() {
ios_base::sync_with_stdio(false); cin.tie(0);
int n, k; cin >> n >> k;
vector<ld> nums(n + 1);
rep(i, 0, n) {
ll x; cin >> x;
nums[i + 1] = x;
}
ld l = -1e13, r = 1e13;
rep(_, 0, 200) {
ld m = (l + r) / 2.00;
ld g;
int opt;
tie(g, opt) = f(nums, m);
if (opt < k)
r = m;
else
l = m;
}
cout << max(0ll, ll(f(nums, l).first + l * ld(k))) << "\n";
}
/*
dp[l - 1] - pf[l - 1] - x
Si voy a cerrar un subarreglo en r, me conviene que sea el que maximice el valor de
dp[l - 1] - pf[l - 1] - x
*/
