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 REP(i,m) for(int i=0; i<m; i++)
#define FOR(i,n,m) for(long long i=n; i<m; i++)
#define INF 1000000001
#define INFL (1LL << 60)
#define MOD 1000000007
#define ALL(v) v.begin(),v.end()
#define pb push_back
#define ll long long int
#define P pair<ll, ll>

int main() {
  int n, k, q;
  cin >> n >> k >> q;
  P a[n];
  REP(i, n){
    int x;
    cin >> x;
    a[i] = {x, i};
  }
  sort(a, a+n);
  bool check[n];
  REP(i, n){
    check[i] = ((n-k-q+i+1 <= a[i].second) && (a[i].second <= k+q-i-2));
  }
  bool isok[n-q+1] = { };
  REP(i, n-q+1){
    REP(j, n){
      if(check[j] && !(i <= j && j < i+q)) isok[i] = true;
    }
  }
  int mm[n-q+1];
  REP(i, n-q+1) mm[i] = a[i+q-1].first - a[i].first;
  int ans = INF;
  REP(i, n-q+1){
    if(!isok[i]) ans = min(ans, mm[i]);
  }
  cout << ans << endl;
  return 0;
}