# 1次元dpで解いてみる
# dp[枚数]での割引最大値

N = int(input())
A, B, C = map(int, input().split())

dp = [0]*(N+1)
for i in range(N+1):
    if i-3 >= 0:
        dp[i] = max(dp[i], dp[i-3]+A)
    if i-5 >= 0:
        dp[i] = max(dp[i], dp[i-5]+B)
    if i-10 >= 0:
        dp[i] = max(dp[i], dp[i-10]+C)

#print(dp)

ans = dp[N]
print(ans)