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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#define F first
#define S second
using namespace std;
using ll = long long;
typedef pair<ll, int> P;
const ll Mod = 1000000007;
const ll INF = Mod * 100LL;
int main() {
  int N,K,Q;
  cin >> N >> K >> Q;
  assert(Q != 1);
  /*
  ll ans = INF;
  vector<ll> A(N);
  vector<P> As(N);
  set<int> Set;
  Set.insert(-1);
  Set.insert(N);
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    As[i].F = A[i];
    As[i].S = i;
  }
  sort(As.begin(),As.end());
  for (int i = 0; i < N; i++) {
    multiset<ll> Sum;
    int p = 0;
    for (auto itr = next(Set.begin(),1); p <= i; itr++) {
      p++;
      auto pitr = prev(itr,1);
      multiset<ll> S;
      for (int j = *pitr+1; j < *itr; j++) {
        S.insert(A[j]);
      }
      
      int len = 0;
      int sz = (int)S.size();
      if (*pitr == -1 && *itr == N) {
        len = max(len,sz);
      } else if (*pitr == -1 || *itr == N) {
        len = max(len,sz - K);
      } else {
        len = max(len,sz - K);
      }
      int cnt = 0;
      for (auto itr = S.begin(); cnt < len; itr++) {
        Sum.insert(*itr);
        cnt++;
      }
    }
    if ((int)Sum.size() >= Q) {
      ans = min(ans,*(next(Sum.begin(),Q-1)) - *(Sum.begin()));
      Set.insert(As[i].S);
    } else {
      break;
    }
  }
  cout << ans << endl;
  */
  return 0;
}