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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
typedef long long LL;
int N,K,Q;
LL A[2001];
typedef pair<LL,int> P;
vector<P>s;
bool used[2001];
int main(){
	cin>>N>>K>>Q;
	REP(i,N){
		cin>>A[i];
		s.push_back(P(A[i],i));
		used[i]=false;
	}
	sort(s.begin(),s.end());
	LL ans=1e18;
	REP(i,N){
		vector<LL>q;
		vector<LL>v;
		REP(j,N){
			if(used[j]==true){
				if(v.size()>=K){
					sort(v.begin(),v.end());
					reverse(v.begin(),v.end());
					for(int k=K-1;k<v.size();k++){
						q.push_back(v[k]);
					}
					v.clear();
				}
			}else{
				v.push_back(A[j]);
			}
		}
		if(v.size()>=K){
			sort(v.begin(),v.end());
			reverse(v.begin(),v.end());
			for(int k=K-1;k<v.size();k++){
				q.push_back(v[k]);
			}
			v.clear();
		}
		
		
		if(q.size()>=Q){
			sort(q.begin(),q.end());
			ans=min(ans,q[Q-1]-q[0]);
		}
		used[s[i].second]=true;
	}
	cout<<ans<<endl;
	
	return(0);
}