Problem:
Score : 300 points

Problem Statement

You have N bamboos. The lengths (in centimeters) of these are l_1, l_2, ..., l_N, respectively.

Your objective is to use some of these bamboos (possibly all) to obtain three bamboos of length A, B, C. For that, you can use the following three kinds of magics any number:

Extension Magic: Consumes 1 MP (magic point). Choose one bamboo and increase its length by 1.

Shortening Magic: Consumes 1 MP. Choose one bamboo of length at least 2 and decrease its length by 1.

Composition Magic: Consumes 10 MP. Choose two bamboos and combine them into one bamboo. The length of this new bamboo is equal to the sum of the lengths of the two bamboos combined. (Afterwards, further magics can be used on this bamboo.)

At least how much MP is needed to achieve the objective?

Constraints

3 \leq N \leq 8

1 \leq C < B < A \leq 1000

1 \leq l_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B C
l_1
l_2
:
l_N

Output

Print the minimum amount of MP needed to achieve the objective.

Sample Input 1

5 100 90 80
98
40
30
21
80

Sample Output 1

23

We are obtaining three bamboos of lengths 100, 90, 80 from five bamboos 98, 40, 30, 21, 80. We already have a bamboo of length 80, and we can obtain bamboos of lengths 100, 90 by using the magics as follows at the total cost of 23 MP, which is optimal.

Use Extension Magic twice on the bamboo of length 98 to obtain a bamboo of length 100. (MP consumed: 2)

Use Composition Magic on the bamboos of lengths 40, 30 to obtain a bamboo of length 70. (MP consumed: 10)

Use Shortening Magic once on the bamboo of length 21 to obtain a bamboo of length 20. (MP consumed: 1)

Use Composition Magic on the bamboo of length 70 obtained in step 2 and the bamboo of length 20 obtained in step 3 to obtain a bamboo of length 90. (MP consumed: 10)

Sample Input 2

8 100 90 80
100
100
90
90
90
80
80
80

Sample Output 2

0

If we already have all bamboos of the desired lengths, the amount of MP needed is 0. As seen here, we do not necessarily need to use all the bamboos.

Sample Input 3

8 1000 800 100
300
333
400
444
500
555
600
666

Sample Output 3

243

Solution in Rust (1.15.1) (judge verdict: Accepted)

// 標準入力から一行を読み取り、指定の型に変換する
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

// 標準入力から一行を読み取り、空白文字で分割し、各要素を指定の型に変換する関数
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace()
        .map(|e| e.parse().ok().unwrap()).collect()
}

// 4つの値の最小値を求める
fn min4(a: i64, b: i64, c: i64, d: i64) -> i64 {
    let mut vec = vec![a, b, c, d];
    vec.sort();

    return vec[0];
}

// 再帰関数(深さ優先探索)
fn dfs(nabc: &Vec<i64>, l: &Vec<i64>, i: usize, a: i64, b: i64, c: i64) -> i64 {
    // 最後の竹を選択済み
    if i == nabc[0] as usize {
        // 一本も選択しないパターンがABCそれぞれに存在する組み合わせは極端に大きな値を返す(除外する)
        if a == 0 || b == 0 || c == 0 { return 10_i64.pow(9); }
        // 絶対値の和がコストになる。後述する理由により、ABCのコストの和から30を差し引いた値を返す.
        return (a - nabc[1]).abs() + (b - nabc[2]).abs() + (c - nabc[3]).abs() - 30;
    }

    // 使用しない場合: ret0
    // 合成した場合: ret1-3
    // 1本目でも合成扱いになるため、最終的にこの分は引いておかなければならない(上で-30する理由).
    let ret0 = dfs(&nabc, &l, i + 1, a, b, c);
    let ret1 = dfs(&nabc, &l, i + 1, a + l[i], b, c) + 10;
    let ret2 = dfs(&nabc, &l, i + 1, a, b + l[i], c) + 10;
    let ret3 = dfs(&nabc, &l, i + 1, a, b, c + l[i]) + 10;

    // 最終的に最も低いコストで収まるパターンを返す.
    return min4(ret0, ret1, ret2, ret3);
}

fn main() {
    let nabc: Vec<i64> = read_vec();
    let mut l: Vec<i64> = Vec::new();

    for _ in 0..nabc[0] as usize
    {
        let li: i64 = read();
        l.push(li);
    }

    // 全探索
    // 各竹には、Aの素材、Bの素材、Cの素材、使わないの4パターン存在するから、これを再帰関数で全探索する.
    // 添字0、ABC全て未選択の状態でスタート
    println!("{}", dfs(&nabc, &l, 0, 0, 0, 0));
}