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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author
 */

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define fr(i,n) for(int i=0;i<(n);++i)
#define Fr(i,n) for(int i=1;i<=(n);++i)
#define ifr(i,n) for(int i=(n)-1;i>=0;--i)
#define iFr(i,n) for(int i=(n);i>0;--i)

int l(vector<int>&a,int m,int k,int q){
    vector<int> v;
    priority_queue<int,vector<int>,greater<int>> pq;
    for(auto&i:a){
        if(i<m){
            sort(v.begin(),v.end());
            fr(j,(int)v.size()-k+1) pq.push(v[j]);
            v.clear();
        }
        else v.emplace_back(i);
    }
    if(pq.size()<q) return INT_MAX;
    fr(i,q-1) pq.pop();
    return pq.top()-m;
}

class ERangeMinimumQueries {
public:
    void solve(istream& in, ostream& out) {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        int n,k,q;
        in>>n>>k>>q;
        vector<int> a(n);
        for(auto&i:a) in>>i;
        a.emplace_back(0);
        int ans=INT_MAX;
        fr(i,n) ans=min(l(a,a[i],k,q),ans);
        out<<ans<<endl;
    }
};

int main() {
    ERangeMinimumQueries solver;
    std::istream& in(std::cin);
    std::ostream& out(std::cout);
    solver.solve(in, out);
    return 0;
}