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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn main() {
    input!(n: usize, k: usize, q: usize, a: [u32; n]);

    let cal = |y: u32| {
        let mut r = 0;
        let mut cand: Vec<u32> = vec![];
        while r < n {
            let mut l = r;
            while l < n && a[l] < y {
                l += 1;
            }
            r = l;
            while r < n && y <= a[r] {
                r += 1;
            }
            if k <= r - l {
                let mut alr = a[l..r].to_vec();
                alr.sort();
                cand.extend(&alr[0..=r - l - k]);
            }
        }
        cand.sort();
        if q <= cand.len() {
            Some(cand[q - 1])
        } else {
            None
        }
    };

    let mut ans = std::u32::MAX;
    for &y in &a {
        if let Some(x) = cal(y) {
            if x - y < ans {
                ans = x - y;
            }
        }
    }
    println!("{}", ans);
}