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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/*
URL_HERE
*/
/*

*/
#ifdef _WIN32
#pragma warning(disable:4996)
#endif
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <unordered_map>
using namespace std;

FILE* _fin = stdin;
FILE* _fout = stdout;
#define PI 3.141592653589793238462643383279502884197169399375105820974
#define ten5p1 100001
#define ten6p1 1000001
#define ten8p1 100000001
#define ten9p1 1000000001
#define uint unsigned int
#define ll long long
#define ull unsigned long long
#define rep(var,n) for(int var=0;var<n;++var)
#define repi(n) rep(i,n)
#define repj(n) rep(j,n)
#define repi1(n) for(int i=1;i<n;++i)
#define repj1(n) for(int j=1;j<n;++j)
int _min(int a, int b) { return a <= b ? a : b; }
int _min(ll a, ll b) { return a <= b ? a : b; }
int _max(int a, int b) { return a >= b ? a : b; }
int _max(ll a, ll b) { return a >= b ? a : b; }
void zero(int* data, int n) { memset(data, 0, sizeof(int) * n); }
void zero(ll* data, int n) { memset(data, 0, sizeof(ll) * n); }
void zero(char* data, int n) { memset(data, 0, sizeof(char) * n); }
char readc() { char var; fscanf(_fin, "%c", &var); return var; }
int readi() { int var; fscanf(_fin, "%d", &var); return var; }
double readlf() { double var; fscanf(_fin, "%lf", &var); return var; }
ll readll() { ll var; fscanf(_fin, "%lld", &var); return var; }
void repread(int* data, int n) { repi(n) data[i] = readi(); }
void repread(ll* data, int n) { repi(n) data[i] = readll(); }
int reads(char* str, int maxsize)
{
	for (;;) { if (fgets(str, maxsize, _fin) == NULL) break; if (str[0] != '\n' && str[0] != '\r') break; }
	int slen = strlen(str); if (slen == 0) return 0;
	if (str[slen - 1] == '\n' || str[slen - 1] == '\r') str[--slen] = 0;
	return slen;
}

#define writec(var) fprintf(_fout,"%c",var)
#define writecsp(var) fprintf(_fout,"%c ",var)
#define writecln(var) fprintf(_fout,"%c\n",var)
#define writei(var) fprintf(_fout,"%d",var)
#define writeisp(var) fprintf(_fout,"%d ",var)
#define writellsp(var) fprintf(_fout,"%lld ",var)
#define writeiln(var) fprintf(_fout,"%d\n",var)
#define writellln(var) fprintf(_fout,"%lld\n",var)
#define writeulln(var) fprintf(_fout,"%llu\n",var)
#define writefln(var) fprintf(_fout,"%f\n",var)
#define writes(str) fprintf(_fout,"%s",str)
#define writesp() fprintf(_fout," ")
#define writeln() fprintf(_fout,"\n")
#define RUN_LOCAL(testfilename) {_fin = fopen(testfilename, "r"); if(_fin==NULL) _fin=stdin;}
#define swap(type,a, b) {type t=a;a=b;b=t;}
#define sort(data,n) std::sort(data,data+n)
//#define mod(a,b) b==0 ? 0 : a==0 ? 0 :(a>0 ? a%b : b + a % b) //what the hell is this...
ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b); }
ll lcm(ll a, ll b) { ll g = gcd(a, b); return (a / g) * b; }
void reverse(char* data, int n) { int k = n >> 1; repi(k) { char tmp = data[i]; data[i] = data[n - i - 1]; data[n - i - 1] = tmp; } }
void reverse(int* data, int n) { int k = n >> 1; repi(k) { int tmp = data[i]; data[i] = data[n - i - 1]; data[n - i - 1] = tmp; } }

#define _Vec(type, name) \
struct name { type* data; int size; int n; };\
void init(name* t, int size) { t->data = (type*)malloc(sizeof(type) * size); t->size = size; t->n = 0; }\
void resize(name* t) { int ns = t->size * 1.2f; t->data = (type*)realloc(t->data, sizeof(type) * ns); t->size = ns; }\
void add(name* t, type val) { if (t->n >= t->size) resize(t); int k = t->n; t->data[k] = val; t->n = k + 1; }\
void free(name* t) { free(t->data);}
_Vec(int, Veci)
_Vec(long long, Vecll)
_Vec(char*, Vecs)

#define _ispal(type, name) int name(type* a, type* b, int n){repi(n){if (a[i] != b[n - i - 1]){return 0;}}return 1;}
_ispal(int, ispali)
_ispal(char, ispalc)

#define _Pair(type, name) struct name{type x, y;}; \
int cmp_##name(const void* _a, const void* _b)\
{\
	name* a = (name*)_a;name* b = (name*)_b;\
	if (a->x < b->x) return -1;\
	else\
	{\
		if (a->x == b->x)\
		{\
			if (a->y == b->y) return 0;\
			if (a->y < b->y) return -1;\
			else return 1;\
		}\
		else\
		{\
			return 1;\
		}\
	}\
}\
void sort_##name(name* d, int n){qsort(d, n, sizeof(name), cmp_##name);}
	_Pair(int, Pairi)

	int cmp_Str(const void* _a, const void* _b) { char* a = *((char**)_a); char* b = *((char**)_b); return strcmp(a, b); }
void sort_Str(char** str, int n) { qsort(str, n, sizeof(char*), cmp_Str); }

ll expmod(ll x, ll n, ll m) {
	ll ans = 1;
	for (; n;) {
		if (n & 1) ans = (ans * x) % m;
		x = (x * x) % m;
		n >>= 1;
	}
	return ans;
}

ll combmod(ll n, ll k, ll m) {
	ll ret = 1;
	ll div = 1;
	for (ll i = 0; i < k; ++i) {
		ret = (ret * (n - i) % m) % m;
		div = (div * (i + 1)) % m;
	}
	div = expmod(div, m - 2, m) % m;
	return (ret * div) % m;
}

int next_perm(int* data, int n)
{
	//find first i,i+1 where d[i]<d[i+1]
	int rootidx = -1;
	for (int i = n - 2; i >= 0; --i)
	{
		if (data[i] < data[i + 1])
		{
			rootidx = i;
			break;
		}
	}

	if (rootidx == -1) return 0;

	int ceilingidx = rootidx + 1;
	//find smallest d[x] after d[rooti] where d[x]>d[rooti]
	for (int i = rootidx + 2; i < n; ++i)
	{
		if (data[i] > data[rootidx] && data[i] < data[ceilingidx])
		{
			ceilingidx = i;
		}
	}

	swap(int, data[rootidx], data[ceilingidx]);
	sort(&data[rootidx + 1], n - rootidx - 1);
	return 1;
}
int count_digits(ll a)
{
	int k = 0;
	for (;;)
	{
		if (a == 0)break;
		k++;
		a /= 10;
	}
	return k;
}

/*
use only when val EXISTS in data
*/
int bs_exists(int* data, int n, int val)
{
	int left = 0;
	int right = n - 1;
	for (; left < right;)
	{
		int mid = (left + right + 1) >> 1;
		if (data[mid] <= val) left = mid;
		else right = mid - 1;
	}
	return right;
}
//------------------------------------------

//------------

void dfs(int* bestmana, int currentmana, int* l, int nl, int maxnl, int a, int b, int c,int tA,int tB,int tC)
{
	if (nl == maxnl)
	{
		//gotta make sure at least 1 stick was added
		if (a == 0 || b == 0 || c == 0)return;
		int totalmanacost = abs(a - tA) + abs(b - tB) + abs(c - tC);//this makes it so that lengths match
		totalmanacost += currentmana;
		if (totalmanacost < *bestmana)
		{
			*bestmana = totalmanacost;
		}
		return;
	}
	
	dfs(bestmana, currentmana, l, nl + 1, maxnl, a , b, c, tA, tB, tC);//dont use stick
	dfs(bestmana, currentmana + 10, l, nl + 1, maxnl, a + l[nl], b, c, tA, tB, tC);//add stick to a
	dfs(bestmana, currentmana + 10, l, nl + 1, maxnl, a, b + l[nl], c, tA, tB, tC);//add stick to b
	dfs(bestmana, currentmana + 10, l, nl + 1, maxnl, a, b, c + l[nl], tA, tB, tC);//add stick to c
}

int main()
{
	
	RUN_LOCAL("xxtop2_0.txt");
	int n = readi();
	int a = readi();
	int b = readi();
	int c = readi();
	static int l[10];
	repread(l, n);

	int bestmana = 1e9;
	dfs(&bestmana, 0, l, 0, n, 0, 0, 0, a, b, c);
	writeiln(bestmana);//initial placement of sticks are not considered joining
	return 0;
}