You are given a positive integer n.
Continuously replace n with the sum of its prime factors.
n multiple times, it should be included in the sum as many times as it divides n.Return the smallest value n will take on.
\n
Example 1:
\n\n\nInput: n = 15\nOutput: 5\nExplanation: Initially, n = 15.\n15 = 3 * 5, so replace n with 3 + 5 = 8.\n8 = 2 * 2 * 2, so replace n with 2 + 2 + 2 = 6.\n6 = 2 * 3, so replace n with 2 + 3 = 5.\n5 is the smallest value n will take on.\n\n\n
Example 2:
\n\n\nInput: n = 3\nOutput: 3\nExplanation: Initially, n = 3.\n3 is the smallest value n will take on.\n\n\n
\n
Constraints:
\n\n2 <= n <= 105给你一个正整数 n 。
请你将 n 的值替换为 n 的 质因数 之和，重复这一过程。
n 能够被某个质因数多次整除，则在求和时，应当包含这个质因数同样次数。返回 n 可以取到的最小值。
\n\n
示例 1：
\n\n输入：n = 15\n输出：5\n解释：最开始，n = 15 。\n15 = 3 * 5 ，所以 n 替换为 3 + 5 = 8 。\n8 = 2 * 2 * 2 ，所以 n 替换为 2 + 2 + 2 = 6 。\n6 = 2 * 3 ，所以 n 替换为 2 + 3 = 5 。\n5 是 n 可以取到的最小值。\n\n\n
示例 2：
\n\n输入：n = 3\n输出：3\n解释：最开始，n = 3 。\n3 是 n 可以取到的最小值。\n\n
\n\n
提示：
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