// Original: https://github.com/tanakh/competitive-rs #[allow(unused_macros)] macro_rules! input { (source = $s:expr, $($r:tt)*) => { let mut iter = $s.split_whitespace(); let mut next = || { iter.next().unwrap() }; input_inner!{next, $($r)*} }; ($($r:tt)*) => { let stdin = std::io::stdin(); let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock())); let mut next = move || -> String{ bytes .by_ref() .map(|r|r.unwrap() as char) .skip_while(|c|c.is_whitespace()) .take_while(|c|!c.is_whitespace()) .collect() }; input_inner!{next, $($r)*} }; } #[allow(unused_macros)] macro_rules! input_inner { ($next:expr) => {}; ($next:expr, ) => {}; ($next:expr, $var:ident : $t:tt $($r:tt)*) => { let $var = read_value!($next, $t); input_inner!{$next $($r)*} }; ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => { let mut $var = read_value!($next, $t); input_inner!{$next $($r)*} }; } #[allow(unused_macros)] macro_rules! read_value { ($next:expr, ( $($t:tt),* )) => { ( $(read_value!($next, $t)),* ) }; ($next:expr, [ $t:tt ; $len:expr ]) => { (0..$len).map(|_| read_value!($next, $t)).collect::>() }; ($next:expr, [ $t:tt ]) => { { let len = read_value!($next, usize); (0..len).map(|_| read_value!($next, $t)).collect::>() } }; ($next:expr, chars) => { read_value!($next, String).chars().collect::>() }; ($next:expr, bytes) => { read_value!($next, String).into_bytes() }; ($next:expr, usize1) => { read_value!($next, usize) - 1 }; ($next:expr, $t:ty) => { $next().parse::<$t>().expect("Parse error") }; } #[allow(dead_code)] fn chmin(x: &mut T, y: T) -> bool where T: PartialOrd + Copy, { *x > y && { *x = y; true } } #[allow(dead_code)] fn chmax(x: &mut T, y: T) -> bool where T: PartialOrd + Copy, { *x < y && { *x = y; true } } #[allow(unused_imports)] use std::cmp::{max, min}; #[allow(unused_imports)] use std::collections::{BTreeMap, BTreeSet, BinaryHeap, VecDeque}; fn main() { input! { n:usize,m:usize, mut ds:[i64;m] } ds.sort(); let mut ans = 1 << 30; for i in 0..m - n + 1 { // 連続したn個のマスを選ぶのが最適 let l = ds[i]; let r = ds[i + n - 1]; let tmp = if l * r > 0 { // 一番左端か右端 max(l.abs(), r.abs()) } else { // 負か正に行った後に折り返すので、重複が発生する。 l.abs() + r.abs() + min(l.abs(), r.abs()) }; chmin(&mut ans, tmp); } println!("{}", ans); }