#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define replr(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define reprl(i, a, b) for (int i = int(a); i >= int(b); --i)
#define rep(i, n) for (int i = 0; i < int(n); ++i)
#define mkp(a, b) make_pair(a, b)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VPI;
typedef vector<VI> VVI;
typedef vector<VVI> VVVI;
typedef vector<VPI> VVPI;
typedef pair<ll, ll> PLL;
typedef vector<ll> VL;
typedef vector<PLL> VPL;
typedef vector<VL> VVL;
typedef vector<VVL> VVVL;
typedef vector<VPL> VVPL;
template<class T> T setmax(T& a, T b) {if (a < b) return a = b; return a;}
template<class T> T setmin(T& a, T b) {if (a < b) return a; return a = b;}
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<typename T>
using indset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#include "boxes.h"
int n, k;
ll l;
ll delivery(int N_ARG, int K_ARG, int L_ARG, int POS_ARG[]) {
n = N_ARG;
k = K_ARG;
l = L_ARG;
VL pos(n);
rep(i, n) pos[i] = POS_ARG[i];
if (k == 1) {
ll ans = 0;
for (ll x : pos) ans += min(x, l-x)*2;
return ans;
}
if (k == n) {
pos.pb(0);
pos.pb(l);
n += 2;
sort(all(pos));
ll ans = l;
rep(i, n-1) {
setmin(ans, 2*pos[i]+2*(l-pos[i+1]));
}
return ans;
}
if (n <= 10) {
VL dp(1<<n, 1e18);
dp[0] = 0;
replr(s, 1, (1<<n)-1) {
/*cout << "SOLVING FOR " << bitset<3>(s) << endl;*/
rep(t, (1<<n)) if ((s | t) == s && __builtin_popcount(s ^ t) <= k) {
VI tmp;
tmp.pb(0);
rep(i, n) if ((s^t)&(1<<i)) tmp.pb(pos[i]);
tmp.pb(l);
ll cost = l;
rep(i, tmp.size()-1) {
setmin(cost, 2*tmp[i]+2*(l-tmp[i+1]));
}
/*cout << " TRYING TO USE " << bitset<3>(t) << " WITH COST " << cost << endl;*/
/*cout << " WITH REMOVED ELEMENTS ";*/
/*replr(i, 1, tmp.size()-2) cout << tmp[i] << " ";*/
/*cout << endl;*/
setmin(dp[s], dp[t] + cost);
}
}
return dp[(1<<n)-1];
}
ll ans = 1e18;
replr(i, 0, n-1) replr(j, i+1, n-1) {
/*if (pos[j] - pos[i] >= (l+1)/2) {*/
ll cur = (j-i-1 + k-1) / k * l;
for (int c = i; c >= 0; c -= k) {
cur += 2*pos[c];
}
for (int c = j; c < n; c += k) {
cur += 2*(l-pos[c]);
}
setmin(ans, cur);
}
return ans;
}
