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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {

	void run() {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		int Q = sc.nextInt();
		int[] A = new int[N];
		int ans = Integer.MAX_VALUE;
		for (int i = 0; i < N; ++i) {
			A[i] = sc.nextInt();
		}
		for (int trial = 0; trial < N; ++trial) {
			PriorityQueue<PriorityQueue<Integer>> pq = new PriorityQueue<>(new Comparator<PriorityQueue<Integer>>() {
				@Override
				public int compare(PriorityQueue<Integer> o1, PriorityQueue<Integer> o2) {
					return Integer.compare(o1.peek(), o2.peek());
				}
			});
			for (int i = 0; i < N; ++i) {
				int t = i;
				PriorityQueue<Integer> tmp = new PriorityQueue<>();
				while (t < N && A[t] >= A[trial]) {
					tmp.add(A[t]);
					++t;
				}
				if (tmp.size() >= K) {
					pq.add(tmp);
				}
				i = t;
			}
			int cnt = 0;
			int max = 0;
			while (cnt < Q && !pq.isEmpty()) {
				PriorityQueue<Integer> cand = pq.poll();
				++cnt;
				max = Math.max(max, cand.poll());
				if (cand.size() >= K)
					pq.add(cand);
			}
			if (cnt == Q) {
				ans = Math.min(ans, max - A[trial]);
			}
		}
		System.out.println(ans);
	}

	public static void main(String[] args) {
		new Main().run();
	}

	void tr(Object... objects) {
		System.out.println(Arrays.deepToString(objects));
	}
}