<p>For an integer array <code>nums</code>, an <strong>inverse pair</strong> is a pair of integers <code>[i, j]</code> where <code>0 &lt;= i &lt; j &lt; nums.length</code> and <code>nums[i] &gt; nums[j]</code>.</p>

<p>Given two integers n and k, return the number of different arrays consisting of numbers from <code>1</code> to <code>n</code> such that there are exactly <code>k</code> <strong>inverse pairs</strong>. Since the answer can be huge, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> n = 3, k = 0
<strong>Output:</strong> 1
<strong>Explanation:</strong> Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pairs.
</pre>

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

<pre>
<strong>Input:</strong> n = 3, k = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>0 &lt;= k &lt;= 1000</code></li>
</ul>

class Solution:
    def kInversePairs(self, n: int, k: int) -> int:
        MOD = 10**9 + 7
        cache = {}

        def inv(n: int, k: int) -> int:
            # Base cases
            if n == 0:
                return 0  # No arrays possible with 0 elements
            if k == 0:
                return 1  # Only one way to arrange n elements with 0 inverse pairs

            # Check if result is already in cache
            key = (n, k)
            if key in cache:
                return cache[key]

            # Calculate the result for (n, k)
            result = inv(n - 1, k)  # Start with arrays of n-1 elements

            # Adjust for cases where nth element creates n inverse pairs
            if k >= n:
                result -= inv(n - 1, k - n)

            result %= MOD  # Apply modulo

            # Add arrays where nth element creates 1 to k inverse pairs
            result = (result + inv(n, k - 1)) % MOD

            # Cache and return the result
            cache[key] = result
            return result

        # Handle the base case
        if k == 0:
            return 1

        # Calculate the number of arrays with exactly k inverse pairs
        # by subtracting arrays with at most k-1 pairs from arrays with at most k pairs
        return (inv(n, k) - inv(n, k - 1)) % MOD
452100