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: Time Limit Exceeded)

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

class Main{
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        int N  =scan.nextInt();
        int K = scan.nextInt();
        int Q = scan.nextInt();
        int[] A = new int[N];
        for(int i=0;i<N;++i)A[i]=scan.nextInt();
        int ans = 1000000001;
        for(int i=0;i<N;++i){
            int left=-1, right=ans;
            while(right-left>1){
                int center=(right+left)/2;
                int cnt=0;
                PriorityQueue<Integer> que = new PriorityQueue<>();
                for(int j=0;j<N;++j){
                    if(A[j]<A[i])que.clear();
                    else{
                        que.add(A[j]);
                        if(que.size()>=K){
                            if(que.poll()-A[i]<=center)++cnt;
                            if(cnt>=Q){
                                break;
                            }
                        }
                    }
                }
                if(cnt>=Q)right=center;
                else left=center;
            }
            ans = Math.min(ans, right);
        }
        System.out.println(ans);


    }
}