<p>Given two <strong>positive</strong> integers <code>n</code> and <code>x</code>.</p>

<p>Return <em>the number of ways </em><code>n</code><em> can be expressed as the sum of the </em><code>x<sup>th</sup></code><em> power of <strong>unique</strong> positive integers, in other words, the number of sets of unique integers </em><code>[n<sub>1</sub>, n<sub>2</sub>, ..., n<sub>k</sub>]</code><em> where </em><code>n = n<sub>1</sub><sup>x</sup> + n<sub>2</sub><sup>x</sup> + ... + n<sub>k</sub><sup>x</sup></code><em>.</em></p>

<p>Since the result can be very large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>

<p>For example, if <code>n = 160</code> and <code>x = 3</code>, one way to express <code>n</code> is <code>n = 2<sup>3</sup> + 3<sup>3</sup> + 5<sup>3</sup></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 10, x = 2
<strong>Output:</strong> 1
<strong>Explanation:</strong> We can express n as the following: n = 3<sup>2</sup> + 1<sup>2</sup> = 10.
It can be shown that it is the only way to express 10 as the sum of the 2<sup>nd</sup> power of unique integers.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 4, x = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> We can express n in the following ways:
- n = 4<sup>1</sup> = 4.
- n = 3<sup>1</sup> + 1<sup>1</sup> = 4.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 300</code></li>
	<li><code>1 &lt;= x &lt;= 5</code></li>
</ul>

# class Solution:
#     def numberOfWays(self, n: int, x: int) -> int:

#         def dp(r, v):
#             if (r, v, x) in memo:
#                 return memo[(r, v, x)]

#             if r == 0:
#                 return 1

#             i, t = v, 0
#             while True:
#                 new = r - i**x
#                 if new < 0:
#                     break

#                 i += 1
#                 t += dp(new, i) % (10**9 + 7)

#             result = t % (10**9 + 7)
#             memo[(r, v, x)] = result
#             return result

#         return dp(n, 1) % (10**9 + 7)
# import math
memo={}
class Solution:
    def numberOfWays(self, n: int, x: int) -> int:
        def numberOfWaysHelp(n,smallest_num):
            if (n,smallest_num,x) in memo:
                return memo[(n,smallest_num,x)]
            if n==0:
                return 1
            out=0
            a=smallest_num
            while True:
                dif = n-a**x
                if dif<0:
                    break
                out+=numberOfWaysHelp(dif,a+1)
                a+=1
            memo[(n,smallest_num,x)]=out
            return out
            
        return numberOfWaysHelp(n,1)%(10**9+7)
494