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 (1.15.1) (judge verdict: Accepted)

#[allow(dead_code)]
fn getline() -> String {
    let mut res = String::new();
    std::io::stdin().read_line(&mut res).ok();
    res
}

#[allow(unused_macros)]
macro_rules! readl {
    ($t: ty) => {
        {
            let s = getline();
            s.trim().parse::<$t>().unwrap()
        }
    };
    ($( $t: ty),+ ) => {
        {
            let s = getline();
            let mut iter = s.trim().split(' ');
            ($(iter.next().unwrap().parse::<$t>().unwrap(),)*)
        }
    };
}

#[allow(unused_macros)]
macro_rules! readlvec {
    ($t: ty) => {
        {
            let s = getline();
            let iter = s.trim().split(' ');
            iter.map(|x| x.parse().unwrap()).collect::<Vec<$t>>()
        }
    }
}

#[allow(unused_macros)]
// macro_rules! debug { ($x: expr) => { println!("{}: {:?}", stringify!($x), $x) } }
macro_rules! debug { ($x: expr) => () }

#[allow(dead_code)]
fn show<T>(iter: T) -> String
where
    T: Iterator,
    T::Item: std::fmt::Display
{
    let mut res = iter.fold(String::new(), |sum, e| sum + &format!("{} ", e));
    res.pop();
    res
}

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
#[allow(unused_imports)]
use std::collections::btree_map::Entry::{Occupied, Vacant};

fn main() {
    use std::io::Write;
    let out = std::io::stdout();
    let mut out = std::io::BufWriter::new(out.lock());
    macro_rules! printf { ($($arg:tt)*) => (write!(out, $($arg)*).unwrap()); }
    macro_rules! printfln { () => (writeln!(out).unwrap()); ($($arg:tt)*) => (writeln!(out, $($arg)*).unwrap()); }
    
    let (n, k, q) = readl!(usize, usize, usize);
    let a = readlvec!(i64);

    let mut b = a.clone();
    b.sort();
    b.dedup();
    let mut fv = vec![false; n];
    let mut ans = 1_000_000_000_000;
    for rm in b {
        debug!(rm);
        let mut ta = Vec::new();
        let mut l = 0;
        let mut r = 0;
        while l < n {
            let mut v = Vec::new();
            while r < n && !fv[r] {
                v.push(a[r]);
                r += 1;
            }
            if k <= v.len() {
                v.sort();
                for &e in v.iter().take(v.len()-k+1) {
                    ta.push(e);
                }
            }
            if l == r {
                l += 1;
                r += 1;
            } else {
                l = r;
            }
        }
        debug!(ta);
        if ta.len() > q-1 {
            ta.sort();
            debug!(ta[q-1]-ta[0]);
            ans = min(ans, ta[q-1]-ta[0]);
        }
        for (i, _) in a.iter().enumerate().filter(|&(_, &e)| e == rm) {
            fv[i] = true;
        }
    }
    printfln!("{}", ans);
}