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)

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
//#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

const int cm = 1 << 17;
char cn[cm], * ci = cn + cm, ct;
inline char getcha() {
	if (ci - cn == cm) { fread_unlocked(cn, 1, cm, stdin); ci = cn; }
	return *ci++;
}
inline int getint() {
	int A = 0;
	if (ci - cn + 16 > cm) while ((ct = getcha()) >= '0') A = A * 10 + ct - '0';
	else while ((ct = *ci++) >= '0') A = A * 10 + ct - '0';
	return A;
}

int main() {
	//cin.tie(0);
	//ios::sync_with_stdio(false);


	int N = getint(), K = getint(), Q = getint();
	ll A[2002];
	int LR[2002];
	rep1(i, N) A[i] = ((ll)getint() << 16) + i;
	rep(i, N + 2) LR[i] = -1;

	priority_queue<int> q;
	priority_queue<int, vector<int>, greater<int>> q2[2000];
	auto q3 = q2;
	int kotae = 1e9;
	int saishou = 2e9;
	sort(A + 1, A + N + 1);
	const int m = (1 << 16) - 1;

	for (int i = N; i >= 1; i--) {
		int a = A[i] >> 16;
		int k = A[i] & m;

		int l = (LR[k - 1] == -1) ? k : LR[k - 1];
		int r = (LR[k + 1] == -1) ? k : LR[k + 1];
		LR[l] = r;
		LR[r] = l;

		if (k + k > l + r) swap(q3[l], q3[k + 1]);
		while (q3[k + 1].size()) {
			int tmp = q3[k + 1].top();
			q3[l].push(tmp);
			q3[k + 1].pop();
		}
		q3[l].push(a);
		while (q3[l].size() >= K) {
			int tmp = q3[l].top();
			q.push(tmp);
			q3[l].pop();
		}
		while (q.size() >= Q) {
			int tmp = q.top();
			chmin(saishou, tmp);
			q.pop();
		}
		chmin(kotae, saishou - a);
	}

	printf("%d", kotae);

	Would you please return 0;
}