#pragma GCC optimize("O2")
#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FOD(i, a, b) for (int i = (a); i >= (b); i--)
#define REP(i, n) for (int i = 0; i < (n); i++)
#define ALL(x) (x).begin(), (x).end()
#define TIME (1.0 * clock() / CLOCKS_PER_SEC)
#define file(name) if (fopen(name".inp", "r")) { freopen(name".inp", "r", stdin); freopen(name".out", "w", stdout); }
using namespace std;
const int MAXN = 3e5 + 5;
const long long INF = 1e18 + 7;
int N, K, a[MAXN];
struct SUB1{
long long dp[2005][2005][2];
void solve(void) {
FOR(i, 0, N) FOR(j, 0, K) dp[i][j][0] = dp[i][j][1] = -INF;
dp[0][0][0] = 0;
FOR(i, 0, N - 1) FOR(j, 0, K) if (dp[i][j][0] != -INF || dp[i][j][1] != INF) {
dp[i + 1][j][0] = max(dp[i][j][0], dp[i][j][1]);
dp[i + 1][j][1] = max(dp[i + 1][j][1], dp[i][j][1] + a[i + 1]);
dp[i + 1][j + 1][1] = max(dp[i + 1][j + 1][1], dp[i][j][0] + a[i + 1]);
}
long long ans = 0;
FOR(i, N, N) FOR(j, 0, K) REP(c, 2) ans = max(ans, dp[i][j][c]);
cout << ans; exit(0);
}
} sub1;
struct SUB2{
#define ii pair <int, int>
#define fi first
#define se second
long long sum[MAXN];
int M = 0, root[MAXN];
ii b[MAXN];
void solve(void) {
FOR(i, 1, N) sum[i] = sum[i - 1] + a[i];
FOR(i, 1, N) {
if (a[i] < 0) continue;
int l = i;
while (l <= N && a[l] >= 0) l++;
b[++M] = {i, l - 1};
i = l - 1;
}
if (M <= K) {
long long ans = 0;
FOR(i, 1, M) ans += sum[b[i].se] - sum[b[i].fi - 1];
cout << ans; exit(0);
}
long long ans = sum[N];
FOR(i, 1, M) ans = max(ans, sum[b[i].se] - sum[b[i].fi - 1]);
cout << ans;
}
} sub2;
void solve(void) {
cin >> N >> K;
FOR(i, 1, N) cin >> a[i];
if (N <= 2000) sub1.solve();
sub2.solve();
}
signed main(void) {
ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
file("FEAST");
//int T; cin >> T; while (T--)
solve();
cerr << "Time elapsed: " << TIME << " s.\n";
return (0 ^ 0);
}
/*
6 2
1 2 3 -10 5 6
*/
