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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
struct Fast {Fast(){std::cin.tie(0);ios::sync_with_stdio(false);cout.precision(20);}} fast;

/* define */
#define FOR(I,X,Y) for(long long (I)=(X);(I)<(Y);(I)++)
#define REP(I,X,Y) for(long long (I)=(Y)-1;(I)>=(X);(I)--)
#define ALL(X) (X).begin(),(X).end()
#define pb push_back
#define COUNT(V,X) (upper_bound((V).begin(),(V).end(),X)-lower_bound((V).begin(),(V).end(),X))
#define debug(x) cerr<<#x<<':'<<x<<endl;
#define DEBUG(v) cerr<<#v<<':';for(auto xXx:v)cerr<<xXx<<' ';cerr<<endl;
#define Yes(X) if(X){cout<<"Yes"<<endl;}else{cout<<"No"<<endl;}
#define YES(X) if(X){cout<<"YES"<<endl;}else{cout<<"NO"<<endl;}
#define ctoi(C) (C-'0')
#define pow2(x) ((long long)((long long)1<<x))

/* alias */
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vii = vector<vector<int>>;
using vl = vector<long long>;
using vll = vector<vector<long long>>;
using pi = pair<int,int>;
using pl = pair<long long,long long>;
template<typename T> using PQ = priority_queue<T>;
template<typename T> using minPQ = priority_queue<T, vector<T>, greater<T>>;

/* const */
const long long dx[] = {1,0,-1,0};
const long long dy[] = {0,1,0,-1};
const long long dx8[] = {1,1,0,-1,-1,-1,0,1};
const long long dy8[] = {0,1,1,1,0,-1,-1,-1};
const long long dx9[] = {1,1,0,-1,-1,-1,0,1,0};
const long long dy9[] = {0,1,1,1,0,-1,-1,-1,0};
const int INF = 1000000007;
const long long LINF = 1000000000000000007;

/* func */
template <typename T> inline bool chmin(T& a, const T& b) {if (a > b) a = b; return a > b;}
template <typename T> inline bool chmax(T& a, const T& b) {if (a < b) a = b; return a < b;}
long long max(long long x,int y){return max(x,(long long)y);}
long long max(int x,long long y){return max((long long)x,y);}
long long min(long long x,int y){return min(x,(long long)y);}
long long min(int x,long long y){return min((long long)x,y);}

/* liblary */

namespace Enumeration{
    //長さNのn進数表記を全列挙する
    vector<vector<long long>> base(long long n,long long N){
        long long SIZE = 1;
        for(long long i = 0;i < N;i++)SIZE *= n;
        vector<vector<long long>> ans(SIZE,vector<long long>());
        for(long long rep = 0;rep < N;rep++){
            SIZE /= n;
            for(long long i = 0;i < ans.size();i++){
                ans[i].push_back((i/SIZE)%n);
            }
        }
        return ans;
    }
    //
}

/* main */

signed main(){
	ll N;
	cin >> N;
	ll A,B,C;
	cin >> A >> B >> C;
	vl l(N);
	FOR(i,0,N){
		cin >> l[i];
	}
	vector<vl> enu = Enumeration::base(4,N);
	ll ans = LINF;
	for(auto v:enu){
		vl a,b,c;
		FOR(i,0,N){
			if(v[i] == 1)a.pb(l[i]);
			if(v[i] == 2)b.pb(l[i]);
			if(v[i] == 3)c.pb(l[i]);
		}
		if(!(a.size()&&b.size()&&c.size()))continue;
		ll tmp = 10*(a.size()+b.size()+c.size()-3);
		tmp += abs( accumulate(ALL(a),0) - A );
		tmp += abs( accumulate(ALL(b),0) - B );
		tmp += abs( accumulate(ALL(c),0) - C );
		chmin(ans,tmp);
	}
	cout << ans << endl;
}