#include <bits/stdc++.h>
using namespace std;
long long delivery(int N, int K, int L, int p[]) {
vector<long long> dpA(N+1, 0);
vector<long long> dpB(N+1, 0);
for(int i = 1; i <= N; i++){
if(i >= K){
dpA[i] = dpA[i-K]+p[i-1]*2;
}else dpA[i] = dpA[0]+p[i-1]*2;
}
priority_queue<pair<long long, int>> pq;
vector<long long> M(N+1, 0);
pq.push({0, N});
for(int i = N-1; i >= 0; i--){
if(i+K > N-1)
dpB[i] = dpB[N] + (L-p[i])*2;
else dpB[i] = dpB[i+K] + (L-p[i])*2;
pq.push({-dpB[i], i});
while(!pq.empty() && pq.top().second > i+K)
pq.pop();
M[i] = -pq.top().first;
}
long long ans = LLONG_MAX;
for(int i = 0; i <= N; i++)
ans = min(ans, dpA[i] + dpB[i]);
for(int i = 0; i < N; i++){
// eliminiamo questo ciclo
ans = min(ans, dpA[i] + M[i] + (long long)L);
/*
for(int j = 1; j <= K; j++){
if(i+j > N) break;
ans = min(ans, dpA[i] + dpB[i+j] + (long long)L);
}
*/
}
// considerando che L è una costante, bisogna solo cercare i e i+j minimi
// posso risolverlo Nlogn
return ans;
}
