def c_synthetic_kadomatsu(N, A, B, C, L):
    from itertools import permutations

    ans = float('inf')
    # 竹の並び方を決める
    for perm in permutations(range(N)):
        # 左から竹をj, k, l本取り、それらを別々に合成し、
        # それらの長さと A, B, C とがどれだけ違うかを計算する
        for j in range(1, N):
            for k in range(1, N - j):
                for l in range(1, N - j - k + 1):
                    cost = 0
                    s, t, u = perm[:j], perm[j:j + k], perm[j + k:j + k + l]
                    a, b, c = [], [], []
                    for _s in s:
                        a.append(L[_s])
                    for _t in t:
                        b.append(L[_t])
                    for _u in u:
                        c.append(L[_u])
                    cost += (len(a) + len(b) + len(c) - 3) * 10
                    cost += abs(sum(a) - A) + abs(sum(b) - B) + abs(sum(c) - C)
                    ans = min(ans, cost)
    return ans

N, A, B, C = [int(i) for i in input().split()]
L = [int(input()) for _ in range(N)]
print(c_synthetic_kadomatsu(N, A, B, C, L))
