#include <bits/stdc++.h>
#include <atcoder/all>

using namespace std;
using namespace atcoder;

using ll = long long;
using mint = modint998244353;
// using mint = modint1000000007;

template <typename T> using vec = vector<T>;
template <typename T> using pa = pair<T, T>;
template <typename T> using mipq = priority_queue<T, vec<T>, greater<T>>;

#define REP(_1, _2, _3, _4, name, ...) name
#define REP1(i, n) for (auto i = decay_t<decltype(n)>{}; (i) < (n); ++(i))
#define REP2(i, l, r) for (auto i = (l); (i) < (r); ++(i))
#define REP3(i, l, r, d) for (auto i = (l); (i) < (r); i += (d))
#define rep(...) REP(__VA_ARGS__, REP3, REP2, REP1)(__VA_ARGS__)
#define rrep(i, r, l) for (int i = (r); i >= (l); --i)
template <typename T> bool chmax(T &a, const T &b) { return (a < b ? a = b, true : false); }
template <typename T> bool chmin(T &a, const T &b) { return (a > b ? a = b, true : false); }

constexpr int INF = 1 << 30;
constexpr ll LINF = 1LL << 60;
constexpr int mov[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};

void solve();

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);
    int T = 1;
    // cin >> T;
    while (T--) solve();
}

// h >= x が達成可能？というのは
// k+1 個に分解して、サイズx以上のピースを2つ作れる？ということ

// iから続けてx以上にするための最小幅、はまあしゃくとりでぜんぶわかる
// 1つ目の切断位置を決め打ちすればまあできるのか...
// 実装でバグらせそー

void solve() {
    int N, L, K;
    cin >> N >> L >> K;
    vec<int> A(N + 2);
    rep(i, N) cin >> A[i + 1];
    A[N + 1] = L;

    int l = 1, r = L;
    while (r - l > 1) {
        int m = (l + r) / 2;

        vec<int> nxt(N + 1, INF), width(N + 1, INF);
        int j = 0;
        rep(i, N + 1) {
            while (j < N + 2 && A[j] - A[i] < m) j++;
            if (j < N + 2) nxt[i] = j, width[i] = j - i;
        }

        vec<int> cum(N + 2, INF);
        rrep(i, N, 0) cum[i] = min(cum[i + 1], width[i]);

        bool ok = false;
        rep(i, N + 1) {
            if (N + 1 + 2 - width[i] - cum[nxt[i]] >= K + 1) {
                l = m;
                ok = true;
                break;
            }
        }
        if (!ok) {
            r = m;
        }
    }

    cout << l << '\n';
}