<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
        if k==0:
            return 1
        dp=[[0]*(k+1) for _ in range(n+1)]
        for i in range(1,n+1):
            dp[i][0]=1
        if n<2:
            return dp[n][k]
        dp[2][1]=1
        for nx in range(3,n+1):
            mx=min(k,(n*(n-1))//2)
            for kx in range(1,mx+1):
                dp[nx][kx]=dp[nx-1][kx]+dp[nx][kx-1]
                if nx<=kx:
                    dp[nx][kx]-=dp[nx-1][kx-nx]
        return dp[n][k]%MOD
167000