/*input
3 1 1
2
11
17
13 3 2
33
66
99
10
83
68
19
83
93
53
15
66
75
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;
typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;
#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"
const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;
template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }
template<typename T>
void print(vector<T> vec, string name = "") {
cout << name;
for (auto u : vec)
cout << u << ' ';
cout << '\n';
}
int N, P, Q;
vi sk;
bool galima(int w) {
if (P + Q >= N) return true;
vii dp(P + 1, vi(Q + 1, 0));
// dp[p][q] = h su p ir q paimta jau [0;h] paimta
for (int p = 0; p <= P; ++p)
{
int jw = 0;
int j2w = 0;
for (int q = 0; q <= Q; ++q)
{
if (dp[p][q] == N) {
if (p + 1 <= P) dp[p + 1][q] = N;
if (q + 1 <= Q) dp[p][q + 1] = N;
continue;
}
int pasiekiu = sk[dp[p][q] + 1] + w;
while (jw + 1 <= N and sk[jw + 1] < pasiekiu)
jw++;
if (p + 1 <= P)
dp[p + 1][q] = max(dp[p + 1][q], jw);
pasiekiu = sk[dp[p][q] + 1] + 2 * w;
while (j2w + 1 <= N and sk[j2w + 1] < pasiekiu)
j2w++;
// printf("p = %d, j2w = %d, q + 1 = %d\n", p, j2w, q + 1);
if (q + 1 <= Q)
dp[p][q + 1] = max(dp[p][q + 1], j2w);
}
}
// for (int p = 0; p <= P; ++p) {
// for (int q = 0; q <= Q; ++q) {
// printf("%d ", dp[p][q]);
// }
// printf("\n");
// }
// printf("w = %d, dp = %d\n", w, dp[P][Q]);
return dp[P][Q] == N;
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cin >> N >> P >> Q;
sk = vi(N + 1);
for (int i = 1; i <= N; ++i)
cin >> sk[i];
sk[0] = 0;
sort(sk.begin() + 1, sk.end());
// for (int i = 0; i <= N; ++i)
// printf("%2d ", sk[i]);
// printf("\n");
int l = 1;
int h = (int)1e9;
int m;
while (l < h) {
m = (l + h) / 2;
if (galima(m))
h = m;
else
l = m + 1;
}
printf("%d\n", l);
}
/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/
