n = int(input())
[[a1, b1], [a2, b2], [a3, b3]] = [list(map(int, input().split())) for _ in range(3)]
assert n > 0 and a1 > 0 and a2 > 0 and a3 > 0 and b1 > 0 and b2 > 0 and b3 > 0
# 最大効率の割引を a1,b1 とする
if a2 * b1 < a1 * b2:
	a1, b1, a2, b2 = a2, b2, a1, b1
if a3 * b1 < a1 * b3:
	a1, b1, a3, b3 = a3, b3, a1, b1
# a1inv : ceil(2^34 / a1)
a1inv = ((2**34 - 1) // a1) + 1
# c1min : a1,b1 の割引の最低使用回数
c1min = max(n // a1 - a2 - a3, 0)
# remain : c1min回を除いた残り枚数
remain = min(a1 * (a2 + a3) + (n % a1), n)
print(
	max(
		# t * a1 < 2^34  -->  floor(ceil(2^34 / a1) * t / 2^34)
		# (t * a1inv >> 34) == (t // a1)
		(t * a1inv >> 34) * b1 + p + q
		for p, s in zip(range(0, a1 * b2, b2), range(remain, -1, -a2))
		for q, t in zip(range(0, a1 * b3, b3), range(s, -1, -a3))
	)
	+ c1min * b1
)