<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>

MOD = 10**9 + 7

class Solution:
    def numberOfWays(self, n: int, x: int) -> int:
        # Calculate the largest integer such that j^x <= n
        def phi(n, x):
            j = 1
            while (j ** x) <= n:
                j += 1
            return j - 1
        
        # Determine the max value of j such that j^x <= n
        max_j = phi(n, x)
        
        # Create dp table: dp[k][j] is the number of ways to sum to k using distinct powers up to j^x
        dp = [[0] * (max_j + 1) for _ in range(n + 1)]
        
        # Base case: dp[0][j] = 1 for all j
        for j in range(max_j + 1):
            dp[0][j] = 1
        
        # Fill the dp table
        for k in range(1, n + 1):  # for each sum we're trying to build
            for j in range(1, max_j + 1):  # for each possible number j^x we can use
                dp[k][j] = dp[k][j-1]  # option to not use j^x
                if k - j**x >= 0:  # option to use j^x if it fits
                    dp[k][j] = (dp[k][j] + dp[k - j**x][j-1]) % MOD
        
        # The answer is the number of ways to express n using distinct powers up to max_j^x
        return dp[n][max_j]
1196