#链家补测# 求元素个数为m，间隔大于等于k的最大子序列和。

#include <bits/stdc++.h>
using namespace std;
int max_sum(vector<int> &A, int n, int m, int k) {
    vector<vector<int>> dp(n, vector<int>(m + 1));
    dp[n - 1][1] = A[n - 1];
    for (int i = n - 2; i >= 0; --i)
        dp[i][1] = max(dp[i + 1][1], A[i]);
    int res = INT_MIN;
    for (int i = n - 1 - k; i >= 0; --i) {
        for (int j = 2; j <= (n - 1 - i) / k + 1; ++j) {
            dp[i][j] = dp[i + k][j - 1] + A[i];
            res = max(res, dp[i][j]);
        }
    }
    return res;
}
int main() {
    int n = 0, m = 0, k = 0;
    cin >> n >> m >> k;
    vector<int> A(n);
    for (int i = 0; i < n; ++i) cin >> A[i];
    int res = max_sum(A, n, m, k);
    cout << res << endl;
}

int max_sum(vector<int> &A, int n, int m, int k) {
    vector<vector<int>> dp(n, vector<int>(m + 1));
    dp[n - 1][1] = A[n - 1];
    for (int i = n - 2; i >= 0; --i) {
        dp[i][1] = max(dp[i + 1][1], A[i]);
    }
    int res = INT_MIN;
    for (int i = n - 1 - k; i >= 0; --i) {
        for (int j = 2; j <= (n - 1 - i) / k + 1; ++j) {
            dp[i][j] = max(dp[i + 1][j], A[i] + dp[i + k][j - 1]);
        }
    }
    return dp[0][m];
}