#include <fstream>

using namespace std;

const int KMAX = 1000 + 5;
const int NMAX = 10000 + 5;

int N;
int v[NMAX];

int dp1[KMAX][NMAX]; //dp1[i][j] = daca iau i subsecvente si ultima se termina pe pozitia j
int dp2[KMAX][NMAX]; //dp2[i][j] = daca iau i subsecvente si ultima se termina pe o pozitie <= j

int main()
{
    ifstream cin("ferma.in");
    ofstream cout("ferma.out");

    //Read
    int K = 0;
    cin >> N >> K;

    for (int i = 1; i <= N; ++ i)
        cin >> v[i];

    //Solve without circularity
    for (int i = 1; i <= K; ++ i) {
        for (int j = 1; j <= N; ++ j) {
            if (j >= 2)
                dp1[i][j] = v[j] + max(dp1[i][j - 1], dp2[i - 1][j - 2]);
            else
                dp1[i][j] = v[j] + dp1[i][j - 1];

            dp2[i][j] = max(dp2[i][j - 1], dp1[i][j]);
        }
    }

    int ans = dp2[K][N];

    //Force first element to be taken
    for (int i = 1; i <= K; ++ i) {
        dp1[1][1] = dp2[1][1] = v[1];
        for (int j = 2; j <= N; ++ j) {
            dp1[i][j] = v[j] + max(dp1[i][j - 1], dp2[i - 1][j - 2]);
            dp2[i][j] = max(dp2[i][j - 1], dp1[i][j]);
        }
    }

    //Tackle circularity
    int sum = 0;
    for (int i = N; i > 1; -- i) {
        sum += v[i];
        ans = max(ans, dp2[K][i - 1] + sum);
    }

    //Print
    cout << ans << '\n';
    return 0;
}