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;
using lint = long long int;
using P = pair<lint, lint>;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()
constexpr int MOD = 1000000007;
constexpr int INF = 2147483647;
void yes(bool expr) {
  cout << (expr ? "Yes" : "No") << "\n";
}
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int N, K, Q;
  cin >> N >> K >> Q;
  vector<int> A(N);
  REP(i, N) cin >> A[i];
  int ans = INF;
  REP(k, N) {
    multiset<int> all;
    multiset<int> tmp;
    REP(i, N) {
      if(A[i] < A[k]) {
        int size = tmp.size() - K + 1;
        REP(j, size) {
          all.insert(*tmp.begin());
          tmp.erase(tmp.begin());
        }
        tmp.clear();
      } else {
        tmp.insert(A[i]);
      }
    }
    int size = tmp.size() - K + 1;
    REP(j, size) {
      all.insert(*tmp.begin());
      tmp.erase(tmp.begin());
    }
    tmp.clear();
    if(all.size() >= Q && *(next(all.begin(), Q-1)) - A[k] < ans) ans = *(next(all.begin(), Q-1)) - A[k];
    all.clear();
  }
  cout << ans << endl;
}