You are given two integer arrays skill and mana. There are n wizards arranged in a fixed order, where wizard i has skill value skill[i], and there are m potions to brew in order, where potion j has mana requirement mana[j].

Brewing a single potion must be done by all wizards sequentially from wizard $0$ to wizard $n - 1$. If wizard i works on potion j, it takes skill[i] * mana[j] units of time.

Each wizard can work on at most one potion at a time, and each potion can be worked on by at most one wizard at a time. A potion must move from wizard i to wizard i + 1 immediately after wizard i finishes, meaning the start times must be chosen so that no potion ever waits between wizards.

Return the minimum possible time at which all m potions have finished brewing (after wizard $n - 1$ completes the last potion).

Constraints:

• n == skill.length

• m == mana.length

• $1 \leq `n`, `m` \leq 5000$

• $1 \leq `mana[i]`, `skill[i]` \leq 5000$

Input: skill = [1, 2]-List[int], mana = [3, 1]-List[int]
Output: 11-int
Explanation: Potion 0 (mana 3) takes 1·3 = 3 with wizard 0 and 2·3 = 6 with wizard 1, so it runs as wizard 0 [0, 3] and wizard 1 [3, 9]. Potion 1 (mana 1) takes 1 and 2 units, and since it must pass to wizard 1 immediately and wizard 1 is busy until time 9, it starts at time 8: wizard 0 [8, 9], wizard 1 [9, 11]. The minimum completion time is 11.

Input: skill = [3]-List[int], mana = [2, 4, 1]-List[int]
Output: 21-int
Explanation: With only one wizard, potions are brewed one after another with no additional constraints, for a total time of 3·2 + 3·4 + 3·1 = 6 + 12 + 3 = 21.

Input: skill = [2, 1, 3]-List[int], mana = [2, 1]-List[int]
Output: 15-int
Explanation: Potion 0 (mana 2) takes [4, 2, 6] across wizards 0–2 and runs as w0 [0, 4], w1 [4, 6], w2 [6, 12]. Potion 1 (mana 1) takes [2, 1, 3] and cannot wait between wizards, so it must reach w2 exactly when w2 is free at time 12; since w0 + w1 take 2 + 1 = 3 units, it starts at time 9: w0 [9, 11], w1 [11, 12], w2 [12, 15]. The final completion time is 15.

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