<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:
        return self.fun1(n, k)

    @functools.lru_cache()
    def fun2(self, n, k):
        nn = n * (n - 1) // 2
        if k < 0:
            return 0
        if k == 0:
            return 1
        if k == nn:
            return 1
        if k > nn:
            return 0

        return (self.fun2(n - 1, k) + self.fun2(n, k - 1) - self.fun2(n - 1, k - n)) % (10**9 + 7)

    def fun1(self, n, k):

        dp = [[0] * (k + 1) for _ in range(n + 1)]

        for i in range(n + 1):
            for j in range(k + 1):
                ii = i * (i - 1) // 2
                if j == 0:
                    dp[i][j] = 1
                elif j == ii:
                    dp[i][j] = 1
                elif j > ii:
                    dp[i][j] = 0
                else:
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
                    if j >= i:
                        dp[i][j] -= dp[i - 1][j - i]
                    dp[i][j] = dp[i][j] % (10**9 + 7)

        return dp[-1][-1]


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