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 Ruby (2.3.3) (judge verdict: Accepted)

#t = Time.now
#N, K, Q = 2000, 10, 10
#as = (0...N).to_a.shuffle
N, K, Q = gets.split.map(&:to_i)
as = gets.split.map(&:to_i)

ia = N.times.map{|i|[i, as[i]]}.sort_by{|arr|arr[1]}

ans = Float::INFINITY
used = Array.new(N) + [true]
(N-Q+1).times do |i|
    if ia[i+Q-1][1] - ia[i][1] >= ans
        used[ia[i][0]] = true
        next
    end
    avail = []
    tmp = []
    d = 0
    (N+1).times do |j|
        if used[j]
            if d >= K
                tmp.sort!
                avail.concat(tmp[0, d-K+1])
            end
            tmp = []
            d = 0
        else
            d += 1
            tmp.push(as[j])
        end
    end
    avail.sort!
    if avail.length >= Q
        cand = avail[Q-1] - ia[i][1]
        ans = cand if ans > cand
    end
    used[ia[i][0]] = true
end

puts ans
#p Time.now - t