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)

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

#define NDEBUG
#ifdef DEBUG
#include "cout11.h"
#undef NDEBUG
#endif
#include <cassert>

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> ii;
typedef pair<ll,ll> llll;
typedef pair<double,double> dd;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<ll> vll;

#define sz(a)  int((a).size())
#define pb  push_back
#define FOR(var,from,to) for(int var=(from);var<=(to);++var)
#define rep(var,n)  for(int var=0;var<(n);++var)
#define rep1(var,n)  for(int var=1;var<=(n);++var)
#define repC2(vari,varj,n)  for(int vari=0;vari<(n)-1;++vari)for(int varj=vari+1;varj<(n);++varj)
#define ALL(c)  (c).begin(),(c).end()
#define RALL(c)  (c).rbegin(),(c).rend()
#define tr(i,c)  for(auto i=(c).begin(); i!=(c).end(); ++i)
#define found(s,e)  ((s).find(e)!=(s).end())
#define mset(arr,val)  memset(arr,val,sizeof(arr))
#define mid(x,y) ((x)+((y)-(x))/2)
#define IN(x,a,b) ((a)<=(x)&&(x)<=(b))

#define INTSPACE 12
char _buf[INTSPACE*1000000 + 3];

int loadint() {
    if (fgets(_buf, INTSPACE+3, stdin)==NULL) return 0;
    return atoi(_buf);
}
int loadvec(vector<int>& v, int N=-1) {
    if (N == -1) {
        N = loadint();
        if (N==0) return 0;
    }
    int bufsize = INTSPACE*N + 3;
    if (fgets(_buf, bufsize, stdin)==NULL) return 0;
    v.resize(N);

    int i=0;
    bool last = false;
    for (char *p=&_buf[0]; ;) {
        char *q = p;
        while (*q > ' ') ++q;
        if (*q == 0x0D || *q == 0x0A) last = true;
        *q = 0;
        v[i++] = atoi(p);
        if (last || i == N) break;
        p = q+1;
    }
    // assert(i <= N);
    return i;
}

int N,K,Q;
vector<int> a;


ll solve() {
#ifdef DEBUG
    printf("N=%d K=%d Q=%d\n", N,K,Q);
#endif
//    vector<ii> u(N);
    map<int,vi> mu;
    rep(i,N) mu[ a[i] ].pb(i);

    vector<ii> ranges;
    ranges.pb(ii(0, N-1)); // inclusive
    int R = ranges.size();
#ifdef DEBUG
    cerr << ranges << endl;
#endif
//    int best = 0x7fffffff;
    vi tmp(ALL(a));
    sort(ALL(tmp));
    int best = tmp[Q-1] - tmp[0];

    for (auto p: mu) {
        // int v = p.first;
#ifdef DEBUG
        cerr << p.first << " " << p.second << endl;
#endif
        int minp = p.first;
        for (int i : p.second) {
            // dividing
            rep(j, R){
                if (IN(i, ranges[j].first, ranges[j].second)) {
                    if (i+1+(K-1) <= ranges[j].second) {
                        ranges.insert(ranges.begin()+j+1, ii(i+1, ranges[j].second));
                    }
                    if (ranges[j].first <= i-1-(K-1)) {
                        ranges[j].second = i-1;
                    } else {
                        ranges.erase(ranges.begin()+j);
                    }
                    break;
                }
            }
            if (ranges.empty()) goto break_loop;

            vi tmp; tmp.reserve(N);
            for (auto r: ranges) {
                for (int c=r.first; c<=r.second; ++c){
                    assert(IN(c, 0, N-1));
                    tmp.pb(a[c]);
                }
            }
            sort(ALL(tmp));
            if (tmp.size()+1 < Q) break;
            // tmp.erase(tmp.begin()+Q);
            int score = tmp[Q-1-1] - minp;
#ifdef DEBUG
            cerr << ranges << tmp << " " << score << endl;
#endif
            best = min(best, score);
        }
    }
break_loop:

    return best;
}


int main() {
    a.reserve(2000);
    loadvec(a,3);
    N=a[0]; K=a[1]; Q=a[2];
    loadvec(a, N);
    cout << solve() << endl;
    return 0;
}