Problem:
Score : 600 points

Problem Statement

You are given an integer sequence A of length N and an integer K.
You will perform the following operation on this sequence Q times:

Choose a contiguous subsequence of length K, then remove the smallest element among the K elements contained in the chosen subsequence (if there are multiple such elements, choose one of them as you like).

Let X and Y be the values of the largest and smallest element removed in the Q operations. You would like X-Y to be as small as possible.
Find the smallest possible value of X-Y when the Q operations are performed optimally.

Constraints

1 \leq N \leq 2000

1 \leq K \leq N

1 \leq Q \leq N-K+1

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K Q
A_1 A_2 ... A_N

Output

Print the smallest possible value of X-Y.

Sample Input 1

5 3 2
4 3 1 5 2

Sample Output 1

1

In the first operation, whichever contiguous subsequence of length 3 we choose, the minimum element in it is 1.
Thus, the first operation removes A_3=1 and now we have A=(4,3,5,2).
In the second operation, it is optimal to choose (A_2,A_3,A_4)=(3,5,2) as the contiguous subsequence of length 3 and remove A_4=2.
In this case, the largest element removed is 2, and the smallest is 1, so their difference is 2-1=1.

Sample Input 2

10 1 6
1 1 2 3 5 8 13 21 34 55

Sample Output 2

7

Sample Input 3

11 7 5
24979445 861648772 623690081 433933447 476190629 262703497 211047202 971407775 628894325 731963982 822804784

Sample Output 3

451211184

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
ll used[10];
ll maxi;

ll dfs(Graph& graph, ll v, ll deep) {
	used[v] = 1;
	vector<ll> maxd;
	maxd.push_back(deep);
	for (int i = 0; i < graph[v].size(); i++) {
		if (used[graph[v][i].to] == 0) {
			maxd.push_back(dfs(graph, graph[v][i].to, deep + 1));
		}
	}
	sort(all(maxd), greater<ll>());
	if (maxd.size() > 2) {
		chmax(maxi, maxd[0] - deep + maxd[1] - deep);
	}
	else {
		chmax(maxi, maxd[0] - deep);
	}
	return maxd[0];
}
*///dfs


#include<iostream>
#include<vector>
#include<set>
#include<queue>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include<cassert>
#include<cmath>
#include<climits>
#include<iomanip>
using namespace std;
#define MOD 1000000007
#define REP(i,n) for(int (i)=0;(i)<(n);(i)++)
#define FOR(i,c) for(decltype((c).begin())i=(c).begin();i!=(c).end();++i)
#define ll long long
#define ull unsigned long long
#define all(hoge) (hoge).begin(),(hoge).end()
typedef pair<ll, ll> P;
const long long INF = 1LL << 60;
typedef vector<ll> Array;
typedef vector<Array> Matrix;


template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}


//sortは初期で昇順 greater<hoge>()で降順
//substr　文字列取り出し
//upper_bound ある値より大きい一番左のイテレータを返す、lowerは以上(setに対して使うとO(N)なので、setのメンバ関数を使う
//stoi


struct Edge {//グラフ
	ll to, cap, rev;
	Edge(ll _to, ll _cap, ll _rev) {
		to = _to; cap = _cap; rev = _rev;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;
void add_edge(Graph& G, ll from, ll to, ll cap, bool revFlag, ll revCap) {//最大フロー求める Ford-fulkerson
	G[from].push_back(Edge(to, cap, (ll)G[to].size()));
	if (revFlag)G[to].push_back(Edge(from, revCap, (ll)G[from].size() - 1));//最小カットの場合逆辺は0にする
}
ll max_flow_dfs(Graph & G, ll v, ll t, ll f, vector<bool> & used)
{
	if (v == t)
		return f;
	used[v] = true;
	for (int i = 0; i < G[v].size(); ++i) {
		Edge& e = G[v][i];
		if (!used[e.to] && e.cap > 0) {
			ll d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
			if (d > 0) {
				e.cap -= d;
				G[e.to][e.rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}
ll max_flow(Graph & G, ll s, ll t)
{
	ll flow = 0;
	for (;;) {
		vector<bool> used(G.size());
		REP(i, used.size())used[i] = false;
		ll f = max_flow_dfs(G, s, t, INF, used);
		if (f == 0) {
			return flow;
		}
		flow += f;
	}
}
void BellmanFord(Graph& G, ll s, Array& d, Array &negative) {//O(|E||V|)
	d.resize(G.size());
	negative.resize(G.size());
	REP(i, d.size())d[i] = INF;
	REP(i, d.size())negative[i] = false;
	d[s] = 0;
	REP(k, G.size() - 2) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
				}
			}
		}
	}
	REP(k, G.size() - 2) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
					negative[G[i][j].to] = true;
				}
				if (negative[i] == true)negative[G[i][j].to] = true;
			}
		}
	}
}
void Dijkstra(Graph& G, ll s, Array& d) {//O(|E|log|V|)
	d.resize(G.size());
	REP(i, d.size())d[i] = INF;
	d[s] = 0;
	priority_queue<P, vector<P>, greater<P>> q;
	q.push(make_pair(0, s));
	while (!q.empty()) {
		P a = q.top();
		q.pop();
		if (d[a.second] < a.first)continue;
		REP(i, G[a.second].size()) {
			Edge e = G[a.second][i];
			if (d[e.to] > d[a.second] + e.cap) {
				d[e.to] = d[a.second] + e.cap;
				q.push(make_pair(d[e.to], e.to));
			}
		}
	}
}
void WarshallFloyd(Graph& G, Matrix& d) {
	d.resize(G.size());
	REP(i, d.size())d[i].resize(G.size());
	REP(i, d.size()) {
		REP(j, d[i].size()) {
			d[i][j] = INF;
		}
	}
	REP(i, G.size()) {
		d[i][i] = 0;
	}
	REP(i, G.size()) {
		REP(j, G[i].size()) {
			d[i][G[i][j].to] = G[i][j].cap;
		}
	}
	REP(i, G.size()) {
		REP(j, G.size()) {
			REP(k, G.size()) {
				chmin(d[j][k], d[j][i] + d[i][k]);
			}
		}
	}
}

class UnionFind {
	vector<int> data;
public:
	UnionFind(int size) : data(size, -1) { }
	bool unionSet(int x, int y) {
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		return x != y;
	}
	bool findSet(int x, int y) {
		return root(x) == root(y);
	}
	int root(int x) {
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	int size(int x) {
		return -data[root(x)];
	}
};

//約数求める //約数
void divisor(ll n, vector<ll>& ret) {
	for (ll i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(ret.begin(), ret.end());

}

//nCrとか
class Combination {
public:
	Array fact;
	Array inv;
	ll mod;
	ll mod_inv(ll x) {
		ll n = mod - 2;
		ll res = 1LL;
		while (n > 0) {
			if (n & 1) res = res * x % mod;
			x = x * x % mod;
			n >>= 1;
		}
		return res;
	}
	ll nCr(ll n, ll r) {
		return ((fact[n] * inv[r] % mod) * inv[n - r]) % mod;
	}
	ll nPr(ll n, ll r) {
		return (fact[n] * inv[n - r]) % mod;
	}
	Combination(ll n, ll _mod) {
		mod = _mod;
		fact.resize(n + 1);
		fact[0] = 1;
		REP(i, n) {
			fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
		}
		inv.resize(n + 1);
		REP(i, n + 1) {
			inv[i] = mod_inv(fact[i]);
		}
	}
};

bool compare_by_b(pair<ll, ll> a, pair<ll, ll> b) {//降順second
	if (a.second != b.second) {
		return a.second > b.second;
	}
	else {
		return a.first > b.first;
	}
}

bool compare_by_a(pair<ll, ll> a, pair<ll, ll> b) {//降順first
	if (a.first != b.first) {
		return a.first > b.first;
	}
	else {
		return a.second > b.second;
	}
}

ll gcd(ll m, ll n) {
	if (n == 0)return m;
	return gcd(n, m % n);
}//gcd

ll lcm(ll m, ll n) {
	return m / gcd(m, n) * n;
}


int main() {
	ll n, k, q;
	cin >> n >> k >> q;
	Array a(n);
	REP(i, n)cin >> a[i];
	ll min = INF;
	REP(i, n) {
		Array temp;
		Array del;
		for (int j =0 ; j < n; j++) {
			if (a[j] >= a[i]&&j!=n-1) {
				temp.push_back(a[j]);
			}
			else {
				if(j==n-1&&a[j]>=a[i])temp.push_back(a[j]);
				if ((ll)temp.size() - k >= 0) {
					sort(all(temp));
					for (int l = 0; l < temp.size() - k + 1; l++) {
						del.push_back(temp[l]);
					}
				}
				temp.clear();
			}
		}
		if (del.size() >= q) {
			sort(all(del));
			chmin(min, del[q - 1] - del[0]);
		}			   
	}
	cout << min;
	return 0;
}