#include <queue>
#include <vector>
#include <cmath>
using namespace std;

typedef long long ll;
const ll INF = 1e18;

int N, M;
ll a[100005];
ll val[400005];
int L[400005], R[400005];
bool del[400005];

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> N >> M;
    for (int i = 1; i <= N; i++) {
        cin >> a[i];
    }

    // 压缩：合并连续同号数为"块"
    vector<ll> blocks;
    ll cur = 0;
    bool has = false;
    for (int i = 1; i <= N; i++) {
        if (a[i] == 0) continue;
        if (!has) {
            cur = a[i];
            has = true;
        } else if ((cur > 0) == (a[i] > 0)) {
            cur += a[i];
        } else {
            blocks.push_back(cur);
            cur = a[i];
        }
    }
    if (has) blocks.push_back(cur);

    // 去掉首尾的负数块
    while (!blocks.empty() && blocks.front() <= 0) blocks.erase(blocks.begin());
    while (!blocks.empty() && blocks.back() <= 0) blocks.pop_back();

    if (blocks.empty()) {
        cout << 0 << endl;
        return 0;
    }

    // 统计正数块数量和总和
    int K = 0;
    ll posSum = 0;
    for (size_t i = 0; i < blocks.size(); i++) {
        if (blocks[i] > 0) {
            K++;
            posSum += blocks[i];
        }
    }

    if (K <= M) {
        cout << posSum << endl;
        return 0;
    }

    // 建双向链表 + 哨兵节点(0和n+1)
    int n = blocks.size();
    priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>> > pq;

    val[0] = INF;
    val[n + 1] = INF;

    for (int i = 1; i <= n; i++) {
        val[i] = abs(blocks[i - 1]);
        del[i] = false;
        L[i] = i - 1;
        R[i] = i + 1;
        pq.push(make_pair(val[i], i));
    }
    del[0] = false;
    del[n + 1] = false;
    R[0] = 1;
    L[n + 1] = n;

    // 反悔贪心：选K-M个代价最小的操作
    int nextNode = n + 2;
    ll totalCost = 0;
    int ops = K - M;

    for (int t = 0; t < ops; t++) {
        while (!pq.empty() && del[pq.top().second]) {
            pq.pop();
        }
        if (pq.empty()) break;

        pair<ll, int> top = pq.top();
        pq.pop();
        ll v = top.first;
        int i = top.second;

        totalCost += v;

        int l = L[i];
        int r = R[i];

        del[l] = true;
        del[i] = true;
        del[r] = true;

        int ll = L[l];
        int rr = R[r];

        // 合并节点：若以后选它 = 反悔选i，改选l和r
        int merged = nextNode++;
        val[merged] = val[l] + val[r] - v;
        del[merged] = false;
        L[merged] = ll;
        R[merged] = rr;
        R[ll] = merged;
        L[rr] = merged;

        pq.push(make_pair(val[merged], merged));
    }

    cout << posSum - totalCost << endl;
    return 0;
}