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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

public static partial class Program
{
    //テンプレここから
    public static int ToInt(this string str) => int.Parse(str);
    public static long ToLong(this string str) => long.Parse(str);
    public static int[] ToInt(this string[] strs) => strs.Select(x => x.ToInt()).ToArray();
    public static long[] ToLong(this string[] strs) => strs.Select(x => x.ToLong()).ToArray();
    public static int[] SpInts(this string str, char opr = ' ') => str.Split(opr).ToInt();
    public static long[] SpLongs(this string str, char opr = ' ') => str.Split(opr).ToLong();
    public static string CRL() => Console.ReadLine();
    public static void CWL(object obj) => Console.WriteLine(obj);
    public static int Diff(int a, int b) => (int)Diff(a, (long)b);
    public static long Diff(long a, long b) => Math.Abs(Math.Max(a, b) - Math.Min(a, b));
    public static int Sign(this int i) => Math.Sign(i);
    public static bool IsEven(this int i) => i % 2 == 0;
    public static int[] dx = new int[] { 0, -1, 0, 1 };
    public static int[] dy = new int[] { -1, 0, 1, 0 };
    public static int Gcd(int a, int b) => (int)Gcd(a, (long)b);
    public static long Gcd(long a, long b)
    {
        if (b == 0)
        {
            return a;
        }

        return Gcd(b, a % b);
    }                       //最大公約数
    public static int Lcm(int a, int b) => (int)Lcm(a, (long)b);
    public static long Lcm(long a, long b) => a / Gcd(a, b) * b;    //最小公倍数
    public static void Swap(ref object a, ref object b)
    {
        var temp = a;
        b = a;
        a = temp;
    }
    //テンプレここまで
}

public static partial class Program
{
    public static void Main()
    {
        var nabc = CRL().SpInts();
        N = nabc[0];
        A = nabc[1];
        B = nabc[2];
        C = nabc[3];
        banboos = new List<int>();
        for (int i = 0; i < nabc[0]; i++)
        {
            banboos.Add(CRL().ToInt());
        }

        CWL(DFS(0, 0, 0, 0));
    }

    public static long N { get; private set; }
    public static long A { get; private set; }
    public static long B { get; private set; }
    public static long C { get; private set; }
    private static List<int> banboos;

    public static long DFS(int cur, long a, long b, long c)
    {
        if (cur == N)
        {
            if (new[] { a, b, c }.Min() < 0)
            {
                return long.MaxValue;
            }
            else
            {
                return Math.Abs(a - A) + Math.Abs(b - B) + Math.Abs(c - C) - 30;
            }
        }

        var ret0 = DFS(cur + 1, a, b, c);                       //その竹を使わない時
        var ret1 = DFS(cur + 1, a + banboos[cur], b, c) + 10;   //
        var ret2 = DFS(cur + 1, a, b + banboos[cur], c) + 10;
        var ret3 = DFS(cur + 1, a, b, c + banboos[cur]) + 10;
        return new[] { ret0, ret1, ret2, ret3 }.Min();
    }
}