We have two arrays arr1 and arr2 which are initially empty. You need to add positive integers to them such that they satisfy all the following conditions:
arr1 contains uniqueCnt1 distinct positive integers, each of which is not divisible by divisor1.arr2 contains uniqueCnt2 distinct positive integers, each of which is not divisible by divisor2.arr1 and arr2.Given divisor1, divisor2, uniqueCnt1, and uniqueCnt2, return the minimum possible maximum integer that can be present in either array.
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Example 1:
\n\n\nInput: divisor1 = 2, divisor2 = 7, uniqueCnt1 = 1, uniqueCnt2 = 3\nOutput: 4\nExplanation: \nWe can distribute the first 4 natural numbers into arr1 and arr2.\narr1 = [1] and arr2 = [2,3,4].\nWe can see that both arrays satisfy all the conditions.\nSince the maximum value is 4, we return it.\n\n\n
Example 2:
\n\n\nInput: divisor1 = 3, divisor2 = 5, uniqueCnt1 = 2, uniqueCnt2 = 1\nOutput: 3\nExplanation: \nHere arr1 = [1,2], and arr2 = [3] satisfy all conditions.\nSince the maximum value is 3, we return it.\n\n
Example 3:
\n\n\nInput: divisor1 = 2, divisor2 = 4, uniqueCnt1 = 8, uniqueCnt2 = 2\nOutput: 15\nExplanation: \nHere, the final possible arrays can be arr1 = [1,3,5,7,9,11,13,15], and arr2 = [2,6].\nIt can be shown that it is not possible to obtain a lower maximum satisfying all conditions. \n\n\n
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Constraints:
\n\n2 <= divisor1, divisor2 <= 1051 <= uniqueCnt1, uniqueCnt2 < 1092 <= uniqueCnt1 + uniqueCnt2 <= 109给你两个数组 arr1 和 arr2 ，它们一开始都是空的。你需要往它们中添加正整数，使它们满足以下条件：
arr1 包含 uniqueCnt1 个 互不相同 的正整数，每个整数都 不能 被 divisor1 整除 。arr2 包含 uniqueCnt2 个 互不相同 的正整数，每个整数都 不能 被 divisor2 整除 。arr1 和 arr2 中的元素 互不相同 。给你 divisor1 ，divisor2 ，uniqueCnt1 和 uniqueCnt2 ，请你返回两个数组中 最大元素 的 最小值 。
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示例 1：
\n\n\n输入：divisor1 = 2, divisor2 = 7, uniqueCnt1 = 1, uniqueCnt2 = 3\n输出：4\n解释：\n我们可以把前 4 个自然数划分到 arr1 和 arr2 中。\narr1 = [1] 和 arr2 = [2,3,4] 。\n可以看出两个数组都满足条件。\n最大值是 4 ，所以返回 4 。\n\n\n
示例 2：
\n\n\n输入：divisor1 = 3, divisor2 = 5, uniqueCnt1 = 2, uniqueCnt2 = 1\n输出：3\n解释：\narr1 = [1,2] 和 arr2 = [3] 满足所有条件。\n最大值是 3 ，所以返回 3 。\n\n
示例 3：
\n\n\n输入：divisor1 = 2, divisor2 = 4, uniqueCnt1 = 8, uniqueCnt2 = 2\n输出：15\n解释：\n最终数组为 arr1 = [1,3,5,7,9,11,13,15] 和 arr2 = [2,6] 。\n上述方案是满足所有条件的最优解。\n\n\n
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提示：
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