# N, M が小さいことに注目して bit DP を行う
# bit DP でもつべき状態は [今の使ったものの集合][現在の重さ] の価値の最大値

# O(2^(N+M) W (N+M)) 間に合うかやや微妙

n, m, w = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
c = list(map(int,input().split()))
d = list(map(int,input().split()))

z = n + m
dp = [[- 10 ** 18] * (w+1) for i in range(1 << z)]

dp[0][0] = 0

for i in range(1 << z):
	for noww in range(w + 1):
		for j in range(z):
			if i >> j & 1:
				if j < n:
					# buy
					if noww - a[j] < 0:
						continue
					dp[i][noww] = max(dp[i][noww], dp[i^(1<<j)][noww-a[j]] + b[j])
				else:
					# use magic
					if noww + c[j-n] > w:
						continue
					dp[i][noww] = max(dp[i][noww], dp[i^(1<<j)][noww+c[j-n]] - d[j-n])

ans = 0
for i in range(1 << z):
	for j in range(w + 1):
		ans = max(ans, dp[i][j])

print(ans)