#include <bits/stdc++.h>
template<class A, class B>
bool maximize(A &a, const B& b) {
if (a < b) {
a = b;
return true;
}
return false;
}
template<class A, class B>
bool minimize(A &a, const B& b) {
if (a > b) {
a = b;
return true;
}
return false;
}
using namespace std;
constexpr int MAX_N = 102, MAX_K = 1E4 + 4;
constexpr long long INF = 0X3F3F3F3F3F3F3F3F, NINF = 0XC0C0C0C0C0C0C0C0;
int N, L, R, K, A[MAX_N];
long long result = INF;
/*
Consider the problem
Give n integers a[1], ..., a[n]
and another n integers b[1], ..., b[n]
Find permutations of (1, ..., n), p and q
such that |a[p[1]] - b[q[1]]| + ... + |a[p[n]] - b[q[n]]|
is minimum
Solution:
Sort a and b
If we want to bring m elements whose indices greater than R into range [L, R]
then we should allocate them in range [R - m + 1, R]
If we want to bring m elements whose indices smaller than L into range [L, R]
then we should allocate them in range [L, L + m - 1]
*/
long long findRightOptimalCost(const int i, const int j, const int k) {
static long long minimum[MAX_N][MAX_N][MAX_K];
static bool visited[MAX_N][MAX_N][MAX_K];
if (k < 0)
return INF;
if (i > R)
return 0;
if (!visited[i][j][k]) {
visited[i][j][k] = true;
minimum[i][j][k] = findRightOptimalCost(i + 1, j, k) + A[i];
if (j <= N) {
const int d = j - i;
minimize(minimum[i][j][k], findRightOptimalCost(i, j + 1, k));
if (k >= d)
minimize(minimum[i][j][k], findRightOptimalCost(i + 1, j + 1, k - d) + A[j]);
}
}
return minimum[i][j][k];
}
long long findLeftOptimalCost(const int i, const int j, const int k) {
static long long minimum[MAX_N][MAX_N][MAX_K];
static bool visited[MAX_N][MAX_N][MAX_K];
if (k < 0)
return INF;
if (i < L)
return 0;
if (!visited[i][j][k]) {
visited[i][j][k] = true;
minimum[i][j][k] = findLeftOptimalCost(i - 1, j, k) + A[i];
if (j >= 1) {
const int d = i - j;
minimize(minimum[i][j][k], findLeftOptimalCost(i, j - 1, k));
if (k >= d)
minimize(minimum[i][j][k], findLeftOptimalCost(i - 1, j - 1, k - d) + A[j]);
}
}
return minimum[i][j][k];
}
signed main() {
#ifdef LOCAL
freopen("input.INP", "r", stdin);
#endif // LOCAL
cin.tie(0) -> sync_with_stdio(0);
cout.tie(0);
cin >> N >> L >> R >> K;
for (int i = 1; i <= N; ++i)
cin >> A[i];
for (int i = L, j, k; i <= R; ++i)
for (j = 0; j < L; ++j)
for (k = 0; k <= K; ++k)
findLeftOptimalCost(i, j, k);
for (int i = R, j, k; i >= L; --i)
for (j = N; j > R; --j)
for (k = 0; k <= K; ++k)
findRightOptimalCost(i, j, k);
minimize(result, min(findLeftOptimalCost(R, L - 1, K), findRightOptimalCost(L, R + 1, K)));
for (int i = L, k; i < R; ++i)
for (k = 0; k <= K; ++k)
minimize(result, findLeftOptimalCost(i, L - 1, k) + findRightOptimalCost(i + 1, R + 1, K - k));
cout << result << '\n';
return 0;
}
