<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 f(self, n, k):
        if k == 0:
            return 1
        if n == 1:
            return 1 if k == 0 else 0
        if (n, k) in self.memo:
            return self.memo[(n, k)]
        p = (n * (n - 1)) // 2
        if k > p:
            return 0
        ans = 0
        upper = min(k, p - k)
        for i in range(upper + 1):
            ans += self.f(n - 1, i)
            ans %= (10**9 + 7)
        self.memo[(n, k)] = ans % (10**9 + 7)
        return self.memo[(n, k)]


    def kInversePairs(self, n: int, k: int) -> int:
        dp = [[0] * (k + 1) for __ in range(n + 1)]
        dp[1][0] = 1
        MOD = 10 ** 9 + 7
        presum = [1] * (k + 1)
        for i in range(2, n + 1):
            cn2 = ((i - 1) * (i - 2)) // 2
            s = 0
            next_presum = []
            for j in range(k + 1):
                low = max(j - i + 1, 0)
                presum_low = 0 if low == 0 else presum[low - 1]
                high = min(cn2, j)
                dp[i][j] = (presum[high] - presum_low + MOD) % MOD
                s += dp[i][j]
                s %= MOD
                next_presum.append(s)
            presum = next_presum
        return dp[n][k]
55100