//#pragma GCC optomize("Ofast")
#pragma GCC optimize("unroll-loops")
//#pragma GCC optimize("O3")
//#pragma GCC target("avx2")
//#pragma GCC target("sse,sse2,sse4.1,sse4.2")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define max_heap priority_queue<pair <ll, pair <ll, ll>>>
#define min_heap priority_queue<pair <ll, ll>, vector<pair <ll, ll>>, greater<pair <ll, ll>>>
//#define min_heap priority_queue<ll, vector<ll>, greater<ll>>
#define sariE cin.tie(NULL); cout.tie(NULL); ios_base::sync_with_stdio(false);
#define filE freopen("in.txt", "r", stdin); freopen("out1.txt", "w", stdout);
#define endl '\n'
#define md(a) (a % mod + mod) % mod
#define pb push_back
//cout << vectorprecision(5) << fixed << f;
//hash prime = 769
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll const maxn = 1e5 + 123;
ll const inf = 2e18;
ll const loG = 23;
ll const mod = 1e9 + 7;
//ll const mod = 998244353;
ll const sq = 400;
ll power(ll a, ll b, ll mod){if(b==0)return 1;if(b==1)return a;ll x = power(a, b / 2, mod);return (((x * x) % mod) * (b % 2 ? a : 1)) % mod;}
int n, A, B, dp[2025][2025];
vector <int> arr, mos;
void Solve(){
cin >> n >> A >> B;
int mn = inf;
for (int i = 1; i < n + 1; i++){
int x; cin >> x;
mos.pb(x);
arr.pb(-x);
mn = min(mn, x);
}
if (A + B >= n){
cout << 1 << endl;
return;
}
sort(arr.begin(), arr.end()); arr.resize(unique(arr.begin(), arr.end()) - arr.begin());
sort(mos.begin(), mos.end()); mos.resize(unique(mos.begin(), mos.end()) - mos.begin());
n = arr.size();
int l = 0, r = mod;
while (r - l > 1){
int mid = (l + r) / 2;
//cout << "mid is " << mid << endl;
for (int i = 1; i < n + 1; i++){
for (int j = 0; j < A + 1; j++){
dp[i][j] = inf;
int x = mos[i - 1];
//cout << "man " << x << endl;
if (x - 2 * mid + 1 > mn){
int y = upper_bound(arr.begin(), arr.end(), -(x - 2 * mid + 1)) - arr.begin();
y = n - y;
dp[i][j] = dp[y][j] + 1;
//cout << "y is " << y << endl;
}
else
dp[i][j] = 1;
if (j){
if (x - mid + 1 > mn){
int y = upper_bound(arr.begin(), arr.end(), -(x - mid + 1)) - arr.begin();
y = n - y;
dp[i][j] = min(dp[i][j], dp[y][j - 1]);
//cout << "yy is " << y << endl;
}
else
dp[i][j] = 0;
}
//cout << i << ' ' << j << ' ' << dp[i][j] << endl;
}
}
if (dp[n][A] <= B)
r = mid;
else
l = mid;
}
cout << r << endl;
}
int main(){
sariE;// filE;
int test = 1;
//cin >> test;
while (test--) Solve();
return 0;
}
