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)

#include <bits/stdc++.h>

#define rep(n) for(int i=0;i<n;i++)
#define repp(j, n) for(int j=0;j<n;j++)
#define reppp(i, m, n) for(int i=m;i<n;i++)
#define all(c) c.begin(), c.end()
#define rall(c) c.rbegin(), c.rend()
#define debug(x) cerr << #x << ": " << x << endl

using namespace std;

typedef long long ll;
typedef pair<ll, ll> Pll;
typedef pair<int, int> Pii;
struct edge{int from, to; ll cost;};

const ll MOD = 1000000007;

int n,k,q;

int main(){
    std::ios::sync_with_stdio(0); cin.tie(0);
    
    cin >> n >> k >> q;
    vector<ll> a(n+1, 0LL);
    rep(n) cin >> a[i];

    ll ans = LLONG_MAX;
    rep(n){
        ll min_ = a[i];
        vector<ll> maxs;
        vector<ll> b;
        repp(j, n+1){
            if(a[j] < min_){
                int m = int(b.size());
                if(m >= k){
                    sort(all(b));
                    repp(j, min(m-k+1, q)) maxs.push_back(b[j]);
                }
                b.clear();
            }else{
                b.push_back(a[j]);
            }
        }

        if(int(maxs.size()) >= q){
            sort(all(maxs));
            ans = min(ans, maxs[q-1]-min_);
        }
    }
    cout << ans << endl;
}