只有一个整数，表示乐曲美妙度的最大值。

【样例说明】
共有5种不同的超级和弦：
音符1 ~ 2，美妙度为3 + 2 = 5
音符2 ~ 3，美妙度为2 + (-6) = -4
音符3 ~ 4，美妙度为(-6) + 8 = 2
音符1 ~ 3，美妙度为3 + 2 + (-6) = -1
音符2 ~ 4，美妙度为2 + (-6) + 8 = 4
最优方案为：乐曲由和弦1，和弦3，和弦5组成，美妙度为5 + 2 + 4 = 11。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#define mp(a,b,c,d) (node){a,b,c,d}
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
using namespace std;
const int maxn = 500005,maxm = 100005,INF = 1000000000;
inline int RD(){int out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();}return out * flag;
}
int N,K,L,R,sum[maxn],mx[maxn][30],Log[maxn];
LL ans = 0;
struct node{int i,l,r,t;};
inline bool operator <(const node& a,const node& b){return sum[a.t] - sum[a.i - 1] < sum[b.t] - sum[b.i - 1];}
priority_queue<node,vector<node> > q;
void RMQ(){Log[0] = -1;for (int i = 1; i <= N; i++) Log[i] = Log[i >> 1] + 1;int t = Log[N];REP(i,N) mx[i][0] = i;for (int i = N; i; i--){for (int j = 1; j <= t; j++){if (i + (1 << j) - 1 > N) break;int t1 = mx[i][j - 1],t2 = mx[i + (1 << j - 1)][j - 1];mx[i][j] = sum[t1] > sum[t2] ? t1 : t2;}}
}
inline int Query(int l,int r){if (l == r) return l;int t = Log[r - l + 1],t1 = mx[l][t],t2 = mx[r - (1 << t) + 1][t];return sum[t1] > sum[t2] ? t1 : t2;
}
void solve(){node u;REP(i,N){if (i + L - 1 > N) break;int r = min(i + R - 1,N);q.push(mp(i,i + L - 1,r,Query(i + L - 1,r)));}for (int i = 1; i <= K; i++){u = q.top(); q.pop();ans += sum[u.t] - sum[u.i - 1];if (u.t - 1 >= u.l) q.push(mp(u.i,u.l,u.t - 1,Query(u.l,u.t - 1)));if (u.t + 1 <= u.r) q.push(mp(u.i,u.t + 1,u.r,Query(u.t + 1,u.r)));}
}
int main(){N = RD(); K = RD(); L = RD(); R = RD();REP(i,N) sum[i] = sum[i - 1] + RD();RMQ();solve();cout<<ans<<endl;return 0;
}