Given two positive integers n and x.
Return the number of ways n can be expressed as the sum of the xth power of unique positive integers, in other words, the number of sets of unique integers [n1, n2, ..., nk] where n = n1x + n2x + ... + nkx.
Since the result can be very large, return it modulo 109 + 7.
For example, if n = 160 and x = 3, one way to express n is n = 23 + 33 + 53.
\n
Example 1:
\n\n\nInput: n = 10, x = 2\nOutput: 1\nExplanation: We can express n as the following: n = 32 + 12 = 10.\nIt can be shown that it is the only way to express 10 as the sum of the 2nd power of unique integers.\n\n\n
Example 2:
\n\n\nInput: n = 4, x = 1\nOutput: 2\nExplanation: We can express n in the following ways:\n- n = 41 = 4.\n- n = 31 + 11 = 4.\n\n\n
\n
Constraints:
\n\n1 <= n <= 3001 <= x <= 5给你两个 正 整数 n 和 x 。
请你返回将 n 表示成一些 互不相同 正整数的 x 次幂之和的方案数。换句话说，你需要返回互不相同整数 [n1, n2, ..., nk] 的集合数目，满足 n = n1x + n2x + ... + nkx 。
由于答案可能非常大，请你将它对 109 + 7 取余后返回。
比方说，n = 160 且 x = 3 ，一个表示 n 的方法是 n = 23 + 33 + 53 。
\n\n
示例 1：
\n\n输入：n = 10, x = 2\n输出：1\n解释：我们可以将 n 表示为：n = 32 + 12 = 10 。\n这是唯一将 10 表达成不同整数 2 次方之和的方案。\n\n\n
示例 2：
\n\n输入：n = 4, x = 1\n输出：2\n解释：我们可以将 n 按以下方案表示：\n- n = 41 = 4 。\n- n = 31 + 11 = 4 。\n\n\n
\n\n
提示：
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