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: Runtime Error)

#include <stdio.h>
#include <bits/stdc++.h>

using namespace std;

typedef double db;
typedef long long ll;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;

typedef pair < db, db > pdd;
typedef pair < db, ld > pdl;
typedef pair < ld, db > pld;
typedef pair < ld, ld > ldp;

typedef pair < ll, ll > pll;
typedef pair < int, ll > pil;
typedef pair < ll, int > pli;
typedef pair < int, int > pii;

#define F first
#define S second

#define en end()
#define bg begin()

#define rev reverse
#define mp make_pair
#define pb push_back

#define y1 y1234567890
#define um unordered_map

#define all(x) x.bg, x.en
#define sz(x) (int)x.size()
#define len(x) (int)strlen(x)

#define sqr(x) ((x + 0ll) * (x))
#define sqrd(x) ((x + 0.0) * (x))

#define forn(i, n) for (int i = 1; i <= n; i++)

const ll inf = (ll)1e18;
const ll mod = (ll)1e9 + 7;

const db eps = (db)1e-9;
const db pi = acos(-1.0);

const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};

const int N = 1005;

int n, k, q, a[N], ar[N], arr[N];

int main() {
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	//freopen(".err", "w", stderr);

	//srand(time(NULL));

	//cin.tie(NULL);
	//cout.tie(NULL);
	//ios_base::sync_with_stdio(false);

	//cout << setprecision(10) << fixed;
	
	cin >> n >> k >> q;

	for (int i = 1; i <= n; i++)
		cin >> a[i];

	int ans = (int)2e9;

	for (int i = 1; i <= n; i++) {
		int m = 0;
		for (int j = 1; j <= n; j++) {
			if (a[j] < a[i])
				continue;
			int ptr = j - 1, sz = 0;
			while (ptr < n && a[ptr + 1] >= a[i])
				ar[++sz] = a[++ptr];
			sort(ar + 1, ar + 1 + sz);
			for (int it = 1; it <= sz - k + 1; it++)
				arr[++m] = ar[it];
			j = ptr;
		}
		sort(arr + 1, arr + 1 + m);
		if (m >= q)
			ans = min(ans, arr[q] - a[i]);
	}

	cout << ans;
	
	//cerr << (clock() + 0.0) / CLOCKS_PER_SEC;

	return 0;
}