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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF ((1LL<<62)-(1LL<<31))
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define bit(n,m)((n>>m)&1)
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}

ll n,k,q,ans=INF;
ll a[2010];
ll temp1[2010];
ll temp2[2010];

int main(){
	scanf("%lld%lld%lld",&n,&k,&q);
	rep(i,0,n)scanf("%lld",a+i);
	a[n]=-1;
	
	while(1){
		ll cnt1=0,cnt2=0;
		rep(i,0,n+1){
			if(a[i]<0){
				if(cnt1>=k){
					sortup(temp1,cnt1);
					rep(j,0,cnt1-(k-1))temp2[cnt2++]=temp1[j];
				}
				cnt1=0;
			}else{
				temp1[cnt1++]=a[i];
			}
		}
		if(cnt2<q)break;
		sortup(temp2,cnt2);
//		rep(i,0,cnt2)printf("%d",temp2[i]);puts("");
		ans=min(ans,temp2[q-1]-temp2[0]);
		rep(i,0,n)if(a[i]==temp2[0])a[i]=-1;
	}
	printf("%lld\n",ans);
	return 0;
}