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 <algorithm>
#include <iostream>
#include <vector>
using namespace std;
const int INF = (int)1e9 + 7;

int main() {
    int n, k; cin >> n >> k;
    int q; cin >> q;
    vector<int> a(n);
    for (auto &ai: a) cin >> ai;
    a.emplace_back(-INF);
    int mi = INF;
    for (int min_i = 0; min_i < n; min_i++) {
        vector<int> cand, tmp;
        for (int i = 0; i < n + 1; i++) {
            if (a[i] < a[min_i]) {
                if ((int)tmp.size() >= k) {
                    sort(begin(tmp), end(tmp));
                    for (int _ = 0; _ < k - 1; _++) tmp.pop_back();
                    cand.insert(cand.end(), tmp.begin(), tmp.end());
                }
                tmp.clear();
            } else {
                tmp.emplace_back(a[i]);
            }
        }
        if ((int)cand.size() >= q) {
            sort(begin(cand), end(cand));
            mi = min(mi, cand[q - 1] - cand[0]);
        }
    }
    cout << mi << endl;
    return 0;
}