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 <bits/stdc++.h>
using namespace std;

#define DUMP(x) cerr << #x << "=" << x << endl
#define DUMP2(x, y) cerr<<"("<<#x<<", "<<#y<<") = ("<<x<<", "<<y<<")"<< endl
#define BINARY(x) static_cast<bitset<16> >(x)

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define REP(i,m,n) for (int i=m;i<(int)(n);i++)

#define in_range(x, y, w, h) (0<=(int)(x) && (int)(x)<(int)(w) && 0<=(int)(y) && (int)(y)<(int)(h))
#define ALL(a) (a).begin(),(a).end()

typedef long long ll;
const int INF   = 1e9;
const ll  INFLL = 1e18;
typedef pair<int, int> PII;
typedef vector<vector<int>> Graph;
int dx[4]={0, -1, 1, 0}, dy[4]={-1, 0, 0, 1};

int main()
{
    ios::sync_with_stdio(false);

    int N, K, Q;
    cin >> N >> K >> Q;
    vector<int> A(N);
    rep(i, N) cin >> A[i];

    int ans = INF;
    rep(i, N) {
        int Y = A[i];
        vector<vector<int>> seqs;
        vector<int> tmp;
        rep(j, N) {
            if (A[j] < Y) {
                if (tmp.size()) {
                    seqs.push_back(tmp);
                    tmp.clear();
                }
            } else {
                tmp.push_back(A[j]);
            }
        }
        if (tmp.size()) {
            seqs.push_back(tmp);
            tmp.clear();
        }

        priority_queue<int, vector<int>, greater<int>> P;
        for (auto seq : seqs) {
            sort(seq.begin(), seq.end());
            for (int j=0; j<(int)seq.size() - K + 1; j++) {
                P.push(seq[j]);
            }
        }

        if (P.size() < Q) continue;
        for (int j=0; j<Q-1; j++) P.pop();
        ans = min(ans, P.top() - Y);
    }
    cout << ans << endl;
}