/* NOI 2019: Feast
Idea: Greedy (combine adjacent values with the same sign and remove negative values from the two ends)
Time Complexity: O(NlogN)
*/
#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
const int MAX_N = 300000 + 5;
const int MAX_L = 20; // ~ Log N
const long long MOD = 1e9 + 7;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> ii;
typedef vector<ii> vii;
typedef vector<vi> vvi;
#define LSOne(S) (S & (-S))
#define isBitSet(S, i) ((S >> i) & 1)
struct block{
int id, l, r;
long long sum;
};
int N, K;
block b[MAX_N];
long long arr[MAX_N];
bool v[MAX_N];
priority_queue<pair<long long, int> > pq;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
cin >> N >> K;
for(int i = 0; i < N; i++) cin >> arr[i];
int j, idx = 0;
for(int i = 0; i < N; i = j) {
b[idx].sum = 0;
b[idx].l = idx - 1;
b[idx].r = idx + 1;
for(j = i; j < N && (arr[i] * arr[j] >= 0); j++) // arr[i] and arr[j] must have the same sign
b[idx].sum += arr[j];
if(!(b[idx].sum < 0 && idx == 0)) idx++; // remove the first block if it is negative
}
if(idx != 0 && b[idx - 1].sum < 0) idx--; // remove the last block if it is negative
if(idx == 0) { // if after remove, there's nothing left
cout << 0;
return 0;
}
b[0].l = b[idx - 1].r = -1; // extreme end points
int remaining = idx / 2 + 1;
for (int i = 0; i < idx; i++)
pq.push({-abs(b[i].sum), i}); // negative of absolute value to have a min priority queue
memset(v, 0, sizeof(v));
while(remaining > K) {
int i = pq.top().second; pq.pop();
if(v[i]) continue;
v[i] = true;
remaining--;
//merge it with left and right
if (b[i].l == -1) { //extreme left
v[b[i].r] = true;
b[b[b[i].r].r].l = -1;
}
else if (b[i].r == -1) { //extreme right
v[b[i].l] = true;
b[b[b[i].l].l].r = -1;
}
else {
v[b[i].r] = true;
v[b[i].l] = true;
b[i].sum += b[b[i].l].sum + b[b[i].r].sum;
if (b[b[i].l].l != -1) {
b[b[b[i].l].l].r = i;
}
b[i].l = b[b[i].l].l;
if (b[b[i].r].r != -1) {
b[b[b[i].r].r].l = i;
}
b[i].r = b[b[i].r].r;
v[i] = 0;
pq.push({-abs(b[i].sum), i});
}
}
long long ans = 0;
for(int i = 0; i < idx; i++)
if (!v[i] && b[i].sum > 0)
ans += b[i].sum;
cout << ans;
}
