//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
template<class T> using PQ = priority_queue<T>;
template<class T> using PQG = priority_queue<T, vector<T>, greater<T> >;
const int INF = 0xccccccc;
const ll LINF = 922337203685477580LL;
template<typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true);}
template<typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true);}
template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second;}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << ' ' << p.second;}

const int N = 3010;

//head

int n, k, m;
int a[N];
ll sum[N], _sum[N];
deque<pair<int, ll>> dq1[N], dq2[N];

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k >> m;
  rep(i, n) cin >> a[i];
  for(int i = n-1; i >= 0; i--) {
    sum[i] = sum[i+1] + a[i];
    _sum[i] = _sum[i+1] - a[i];
  }

  dq1[0].emplace_back(-1, sum[0]);
  dq2[0].emplace_back(-1, _sum[0]);
  rep(i, n) {
    for(int j = k-1; j >= 0; j--) {
      while(!dq1[j].empty() and dq1[j].front().first < i-m) dq1[j].pop_front();
      while(!dq2[j].empty() and dq2[j].front().first < i-m) dq2[j].pop_front();
      ll now = 0;
      if(!dq1[j].empty()) chmax(now, dq1[j].front().second - sum[i+1]);
      if(!dq2[j].empty()) chmax(now, dq2[j].front().second - _sum[i+1]);
      int n1 = now+sum[i+1];
      int n2 = now+_sum[i+1];
      while(!dq1[j+1].empty() and dq1[j+1].back().second <= n1) dq1[j+1].pop_back();
      while(!dq2[j+1].empty() and dq2[j+1].back().second <= n2) dq2[j+1].pop_back();
      dq1[j+1].emplace_back(i, n1);
      dq2[j+1].emplace_back(i, n2);
    }
  }
  cout << max(dq1[k].back().second, dq2[k].back().second) << endl;
}