#pragma GCC optimize("Ofast,unroll-loops,inline,fast-math,omit-frame-pointer")
#pragma GCC optimize("no-stack-protector,no-trapping-math,rename-registers")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt,tune=native")
#pragma GCC optimize("tree-vectorize")
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define vll vector<ll>
#define len(x) (ll)x.size()
const ll inf = 1e9, infl = 1e18;
const ll MOD = 1e9 + 7;
const ll maxn = 2e5 + 5;
ll n, m, k = 0;
ll a[105], memo[5005][105];
void _() {
// r = n subtasklari
// Observation: eger range icinden cixaracagin ve range icine getireceyin indexleri secsen bes edir
// cunki cost onlarin yerini nece deyissen de eyni qalir (eger rangeden eyni terefdedise)
// n <= 13 subtaski ucun onda sadece range icinden max k dene element secsek colden optimal olan k dene
// elementi secmek ucun bitmask ata bilerik cunki en pis halda n / 2 dene element secmeliyik range colunden.
// Cunki, (k <= min(length of range, n - length of range)) max qiymeti ucun length of range = n / 2,
// O(2^12) = O(4096), o da rahatliqla kecir.
// Rahatliqla olmasa da kecdi(Sliding Windows)
// n <= 50 subtaski ucun bitmask ata bilmerem, dp elemeliyem necese
// Observation: Max lazim olan cost n <= 100 ucun bele 5000-den cox deyil
ll l, r;
cin >> n >> l >> r >> m;
m = min(m, 5000ll);
l --, r --;
for(ll i = 0; i < n; i ++){
cin >> a[i];
}
ll sz = n / 2, sz_ = n - n / 2, le = r - l + 1, cv = inf;
for(ll cs = 0; cs <= m; cs ++){
for(ll ct = 0; ct <= le; ct ++){
memo[cs][ct] = inf;
}
}
{
ll cm = 0;
for(ll i = l; i <= r; i ++) cm += a[i];
memo[0][le] = cm;
}
for(ll msk = 0; msk < (1ll << sz); msk ++){
if(__builtin_popcount(msk) > le) continue;
ll cost = 0, sum = 0, ind = l, x = msk;
while(x){
ll nd = __builtin_ctz(x);
sum += a[nd];
cost += abs(nd - ind);
ind ++;
x &= x - 1;
}
// for(ll i = 0; i < sz; i ++){
// if(msk & (1ll << i)){
// sum += a[i];
// cost += abs(i - ind);
// if(cost > m) break;
// ind ++;
// }
// }
if(cost <= m){
memo[cost][__builtin_popcount(msk)] = min(memo[cost][__builtin_popcount(msk)], sum);
}
}
for(ll cs = 1; cs <= m; cs ++){
for(ll ct = 0; ct <= le; ct ++){
memo[cs][ct] = min(memo[cs][ct], memo[cs - 1][ct]);
}
}
for(ll msk_ = 0; msk_ < (1ll << sz_); msk_ ++){
if(__builtin_popcount(msk_) > le) continue;
ll cost = 0, sum = 0, ind = r - __builtin_popcount(msk_) + 1, x_ = msk_;
while(x_){
ll nd = __builtin_ctz(x_);
sum += a[nd + sz];
cost += abs((nd + sz) - ind);
ind ++;
x_ &= x_ - 1;
}
// for(ll i_ = 0; i_ < sz_; i_ ++){
// if(msk_ & (1ll << i_)){
// sum += a[i_ + sz];
// cost += abs((i_ + sz) - ind);
// if(cost > m) break;
// ind ++;
// }
// }
if(cost <= m){
cv = min(cv, memo[m - cost][le - __builtin_popcount(msk_)] + sum);
}
}
cout << cv << '\n';
}
signed main() {
cin.tie(0)->sync_with_stdio(0);
ll t = 1;
// cin >> t;
while(t --) _();
}
