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 <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <cstring>
#include <map>
#include <queue>
#include <cmath>
#include <complex> // complex<double> a(1.2 , 2.3);// real(): 1.2, imag()2.3
using namespace std;

#define MOD 1000000007
#define ll long long
#define ld long double
#define FOR(i,a,b) for(ll i=(ll)a;i<(ll)b;i++)
#define rep(i,n) FOR(i,0,n)
#define pb push_back
#define mp make_pair
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define rmsame(a) sort(all(a)),a.erase(unique(all(a)), a.end())
#define rmvector(a,b) rep(i,a.size())rep(j,b.size())if(a[i]==b[j]){a.erase(a.begin()+i);i--;break;}
#define pq_pair_tB priority_queue <pair<ll,ll>, vector<pair<ll,ll> > , greater<pair<ll,ll> > > 
#define pq_pair_ts priority_queue <pair<ll,ll> > //第二成分の大きさが関係ない
template<typename X> bool exist(vector<X> vec, X item){return find(all(vec), item)!=vec.end();}
ll gcd(ll a, ll b){if(b==0)return a;return gcd(b,a%b);}
ll lcm(ll a, ll b){return a/gcd(a,b)*b;}

ll N , K , Q;
vector<ll> A ;
vector<ll> lst;

bool check(ll value){
    ll cnt = 0;
    ll ret = 0;
    
    rep(i,N){
        if(A[i] >= value){
            cnt++;
        }
        else{
            if(cnt >= K) ret += (cnt-K+1);
            cnt = 0;
        }
    }

    if(cnt >= K) ret += (cnt-K+1);
    cnt  = 0 ;
    if(ret >= Q) return true;
    else return false;
}



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

    cin >> N >> K >> Q;
    rep(i,N){
        ll inp;
        cin >> inp;
        A.pb(inp);
        lst.pb(inp);
    }
    sort(all(lst));
    
    if(Q == 1){ cout << 0 << endl; return 0;}
    
    ll l = 0;
    ll r = lst.size();
 
    
    while(l + 1 < r){
        ll m = (l+r)/2;
        ll value = lst[m];
        if(check(value)) {
            if(m)
            l = m;
        }
        else r = m;
    }
    //取り除ける　最小値　の最大の可能性
    //cerr << lst[l] << " " << lst[r]  << endl;
    
    ll ans = 10000000000;

    rep(i, l+1){
        if((i+Q-1) > N-1) continue;
        cerr << lst[i+Q-1] << " " << lst[i] << endl;
        ans = min(lst[i+Q-1] - lst[i], ans);
    }
    cout << ans << endl;
    
    //cout << fixed << setprecision(16) << ans << endl;
    return 0;
}