<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:
        dp = [[0 for i in range(k+1)] for j in range(n+1)]
        dp[0][0] = 1
        presum = [0 for i in range(k+2)]
        for i in range(k+1):
            presum[i+1] = presum[i] + dp[0][i]
        M = int(1e9+7)
        for i in range(1, n+1):
            tmp = [0 for i in range(k+2)]
            for j in range(k+1):
                # dp[i][j] = \sum dp[i-1][m] for m in [j, j-(i-1)]
                # for m in range(j, max(j-i, -1), -1):
                #     dp[i][j] += dp[i-1][m]
                dp[i][j] = presum[j+1] - presum[max(0, j-i+1)]
                dp[i][j] %= M
                tmp[j+1] = tmp[j] + dp[i][j]
                # print(i,j,dp[i][j])
            presum = [x % M for x in tmp]
        return dp[n][k]
55900