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: Wrong Answer)

// E.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

int main(int argc, char *argv[])
{
	int n, k, q;
	cin >> n >> k >> q;
	vector<int> a(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	int ans = 1 << 30;
	for (int i = 0; i < n; ++i) {
		vector<int> x, y;
		for (int j = 0; j <= n; ++j) {
			if (j == n || a[j] < a[i]) {
				if (y.size() >= k) {
					sort(y.begin(), y.end());
					x.insert(x.end(), y.begin(), y.end() - k + 1);
				}
				y.clear();
			} else {
				y.push_back(a[j]);
			}
		}
		if (x.size() < q) {
			break;
		}
		sort(x.begin(), x.end());
		ans = min(ans, x[q - 1] - x[0]);
	}
	cout << ans << endl;
}