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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
  
#define ll long long
#define all(aaa) aaa.begin(), aaa.end()

const int INF = 1e9 + 5;
int n, k, q, ans = INF;

void rec(vector<int> v, int mn = INF, int mx = -1, int cq = 0) {
    if (cq == q) {
        ans = min(ans, mx - mn);
    }
    else {
        for (int i = 0; i <= v.size() - k; i++) {
            vector<int> nv = v;

            int x = i;
            for (int j = 0; j < k; j++) {
                if (nv[x] > nv[i + j])
                    x = i + j;                
            }

            int nmx = max(mx, nv[x]), 
                nmn = min(mn, nv[x]);

            nv.erase(nv.begin() + x);

            rec(nv, nmn, nmx, cq + 1);
        }
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n >> k >> q;

    vector<int> v;

    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        v.push_back(x);
    }

    rec(v);

    cout << ans;
 
    return 0;   
}