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 <algorithm>
#include <iostream>
#include <iomanip>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <numeric>
#include <bitset>

using namespace std;
using ll = long long;
using ld = long double;
const ll mod =  1000000007;

#define REP(i, n) for (ll i = 0; i < (n); ++i)
#define REP_FROM(i, j, n) for (ll i = (j); i < (n); ++i)
#define all(x) (x).begin(),(x).end()

ll power(ll base, ll exponent, ll module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ ll root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
ll inverse(ll x){ return power(x, mod - 2, mod);}
ll gcd(ll a, ll b) {if(a < b) gcd(b, a);ll r;while(r=a%b) {a=b;b=r;}return b;}
struct combination{ vector<ll> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2, mod); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } ll C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };

int bsearch(const vector<ll>& a, ll x){
  int lft = 0;
  int rgt = a.size();
  while (rgt - lft > 1) {
    int mid = (lft + rgt) / 2;
    if(a[mid] < x) {
      lft = mid;
    }
    else {
      rgt = mid;
    }
  }
  return lft;
}

using PQ = priority_queue<ll, vector<ll>, greater<ll>>;

signed main() {
  ios::sync_with_stdio(false);
  ll n, k, q;
  cin >> n >> k >> q;
  vector<ll> a(n);
  vector<ll> b(n);
  REP(i, n) cin >> a[i];
  REP(i, n) b[i] = a[i];
  PQ p;
  vector<ll> s;
  sort(all(b));
  ll ans = 1e9;
  REP(i, n - k + 1) {
    if(i > 0 && b[i-1] == b[i]) continue;
    p = PQ();
    s.clear();
    REP(j, n) {
      if(a[j] < b[i]) {
        while(p.size() >= k) {
          s.push_back(p.top());
          p.pop();
        }
        p = PQ();
      }
      else {
        p.push(a[j]);
      }
    }
    while(p.size() >= k) {
      s.push_back(p.top());
      p.pop();
    }
    if(s.size() < q) continue;
    sort(all(s));
    ans = min(ans, s[q-1] - s[0]);
  }
  cout << ans << endl;
  return 0;
}