// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



#define rep(i, n) for (int i = 0; i < n; ++i)

#define ALL(a) (a).begin(), (a).end()



template <class T>

inline bool chmin(T &a, T b)

{

  if (a > b)

  {

    a = b;

    return true;

  }

  return false;

}

template <class T>

inline bool chmax(T &a, T b)

{

  if (a < b)

  {

    a = b;

    return true;

  }

  return false;

}



using ll = long long;

using ld = long double;

using P = pair<int, int>;

using DoP = pair<double, double>;

using edge = struct { int to; ll cost; };



const int di[] = {0, 1, 0, -1, 1, 1, -1, -1};

const int dj[] = {1, 0, -1, 0, 1, -1, 1, -1};

const int INF = 1e9 + 100;

const ll INF64 = 1ll << 40;

const int mod = 1000000007;



struct Sieve

{

  int n;

  vector<int> f, primes;

  Sieve(int n = 1) : n(n), f(n + 1)

  {

    f[0] = f[1] = -1;

    for (ll i = 2; i <= n; ++i)

    {

      if (f[i])

        continue;

      primes.push_back(i);

      f[i] = i;

      for (ll j = i * i; j <= n; j += i)

      {

        if (!f[j])

          f[j] = i;

      }

    }

  }

  bool isPrime(int x) { return f[x] == x; }

  vector<int> factorList(int x)

  {

    vector<int> res;

    while (x != 1)

    {

      res.push_back(f[x]);

      x /= f[x];

    }

    return res;

  }

  vector<P> factor(int x)

  {

    vector<int> fl = factorList(x);

    if (fl.size() == 0)

      return {};

    vector<P> res(1, P(fl[0], 0));

    for (int p : fl)

    {

      if (res.back().first == p)

      {

        res.back().second++;

      }

      else

      {

        res.emplace_back(p, 1);

      }

    }

    return res;

  }

};

struct UnionFind

{

  vector<int> par;

  vector<int> siz;

  UnionFind(int N) : par(N), siz(N, 1) {for (int i = 0; i < N; i++) par[i] = i;}

  

  void init(int N) {

    siz.assign(N, 1);

    par.resize(N);

    for(int i = 0; i < N; ++i) par[i] = i;

  }



  int root(int x) {

    while(par[x] != x) {x = par[x] = par[par[x]];}

    return x;

  }



  bool unite(int x, int y)

  {

    x = root(x);

    y = root(y);

    if(x == y) return false;



    if(siz[x] < siz[y]) swap(x, y);

    siz[x] += siz[y];

    par[y] = x;

    return true;



  }



  bool isSame(int x, int y) {return root(x) == root(y); }



  int size(int x) { return siz[root(x)]; }

};

struct mint

{

  ll x; // typedef long long ll;

  mint(ll x = 0) : x((x % mod + mod) % mod) {}

  mint operator-() const { return mint(-x); }

  mint &operator+=(const mint a)

  {

    if ((x += a.x) >= mod)

      x -= mod;

    return *this;

  }

  mint &operator-=(const mint a)

  {

    if ((x += mod - a.x) >= mod)

      x -= mod;

    return *this;

  }

  mint &operator*=(const mint a)

  {

    (x *= a.x) %= mod;

    return *this;

  }

  mint operator+(const mint a) const

  {

    mint res(*this);

    return res += a;

  }

  mint operator-(const mint a) const

  {

    mint res(*this);

    return res -= a;

  }

  mint operator*(const mint a) const

  {

    mint res(*this);

    return res *= a;

  }

  mint pow(ll t) const

  {

    if (!t)

      return 1;

    mint a = pow(t >> 1);

    a *= a;

    if (t & 1)

      a *= *this;

    return a;

  }



  // for prime mod

  mint inv() const

  {

    return pow(mod - 2);

  }

  mint &operator/=(const mint a)

  {

    return (*this) *= a.inv();

  }

  mint operator/(const mint a) const

  {

    mint res(*this);

    return res /= a;

  }

};

// h(n, r) = c(n+r-1, r)

struct combination

{

  vector<mint> fact, ifact;

  combination(int n) : fact(n + 1), ifact(n + 1)

  {

    assert(n < mod);

    fact[0] = 1;

    for (int i = 1; i <= n; ++i)

      fact[i] = fact[i - 1] * i;

    ifact[n] = fact[n].inv();

    for (int i = n; i >= 1; --i)

      ifact[i - 1] = ifact[i] * i;

  }

  mint operator()(int n, int k)

  {

    if (k < 0 || k > n)

      return 0;

    return fact[n] * ifact[k] * ifact[n - k];

  }

};







int main(){

  int n, a, b, c;

  cin >> n >> a >> b >> c;

  vector<int> l(n);

  rep(i, n) cin >> l[i];

  int ans = INF;

  for(int bit = 0; bit < 1<<(2*n); ++bit) {

    vector<int> aa, bb, cc;

    for(int i = 0; i < 2*n; i+=2) {

      bool ll = (bit & 1<<i), rr = (bit & 1<<(i+1));

      if(!ll && !rr) continue;

      else if(!ll && rr) aa.push_back(l[i/2]);

      else if(ll && !rr) bb.push_back(l[i/2]);

      else cc.push_back(l[i/2]);

    }

    if(!aa.size() || !bb.size() || !cc.size()) continue;

    int cost = 10*(aa.size()+bb.size()+cc.size()-3);

    cost += abs(a - accumulate(ALL(aa), 0));

    cost += abs(b - accumulate(ALL(bb), 0));

    cost += abs(c - accumulate(ALL(cc), 0));



    ans = min(ans, cost);

  }

  cout << ans << endl;

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;



// エイリアス

using  ll = long signed long;

using ull = long unsigned long;

using  ld = long double;

using   P = pair<int, int>;

using llP = pair<ll, ll>;

using DoP = pair<double, double>;



// 汎用マクロ

#define ALL(a) (a).begin(), (a).end()

#define RALL(a) (a).rbegin(), (a).rend()

#define rep(i, n) for (int i = 0; i < n; ++i)

#define FOREACH(i,q) for (auto &i : q)

#define UNIQUE(v) do { sort((v).begin(), (v).end()); (v).erase(unique((v).begin(), (v).end()), (v).end()); } while (false)

#define SZ(x) ((int)(x).size())

template <class T>inline bool chmin(T &a, const T b) {if (a > b) { a = b; return true;} return false;}

template <class T>inline bool chmax(T &a, const T b) {if (a < b) { a = b; return true;} return false;}

template <class T>inline void line_out(const vector<T> vec) {int n = SZ(vec); rep(i, n) { cout << vec[i]; if(i < n-1) cout << " ";}cout << endl;}

const int di[] = {0, 1, 0, -1, 1, 1, -1, -1};

const int dj[] = {1, 0, -1, 0, 1, -1, 1, -1};

const int INF = 1 << 28;

const ll INF64 = 1ll << 60;

const int mod = 1000000007;

//const int mod = 998244353;

struct mint

{

  ll x; // typedef long long ll;

  mint(ll x = 0) : x((x % mod + mod) % mod) {}

  mint operator-() const { return mint(-x); }

  mint &operator+=(const mint a)

  {

    if ((x += a.x) >= mod)

      x -= mod;

    return *this;

  }

  mint &operator-=(const mint a)

  {

    if ((x += mod - a.x) >= mod)

      x -= mod;

    return *this;

  }

  mint &operator*=(const mint a)

  {

    (x *= a.x) %= mod;

    return *this;

  }

  mint operator+(const mint a) const

  {

    mint res(*this);

    return res += a;

  }

  mint operator-(const mint a) const

  {

    mint res(*this);

    return res -= a;

  }

  mint &operator++()

  {

    if ((x += 1) >= mod)

      x -= mod;

    return *this;

  }

  mint &operator--()

  {

    if ((x += mod - 1) >= mod)

      x -= mod;

    return *this;

  }

  mint operator*(const mint a) const

  {

    mint res(*this);

    return res *= a;

  }

  mint pow(ll t) const

  {

    if (!t)

      return 1;

    mint a = pow(t >> 1);

    a *= a;

    if (t & 1)

      a *= *this;

    return a;

  }



  // for prime mod

  mint inv() const

  {

    return pow(mod - 2);

  }

  mint &operator/=(const mint a)

  {

    return (*this) *= a.inv();

  }

  mint operator/(const mint a) const

  {

    mint res(*this);

    return res /= a;

  }

};



int main()

{

  int n, a, b, c;

  cin >> n >> a >> b >> c;

  vector<int> l(n);

  rep(i, n) cin >> l[i];

  

  int ans = INF;



  for(int bit = 0; bit < 1<<(2*n); ++bit) {

    int aa = 0, bb = 0, cc = 0;

    int cost = 0;



    rep(i, n) {

      bool f1 = bit&(1<<(2*i)), f2 = bit&(1<<(2*i+1));

      if(!f1 && !f2) continue;

      cost += 10;

      if(f1 && !f2) aa += l[i];

      if(!f1 && f2) bb += l[i];

      if(f1 && f2) cc += l[i];

    }



    //cout << aa << " " << bb << " " << cc << endl;

    if(aa == 0 || bb == 0 || cc == 0) continue;

    cost += abs(aa-a) + abs(bb-b) + abs(cc-c);

    cost -= 30;

    chmin(ans, cost);

  }

  cout << ans << endl;

}


// Fastest known solution for this problem: