<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>

using i32 = int;
using i64 = long long;
const i32 MOD = 1e9 + 7;
class Solution {
public:
    int kInversePairs(int n, int k) {
      vector <vector <i64>> dp (n, vector <i64> (k + 1));
      dp[0][0] = 1;
      for (i32 i = 1; i <= n; ++i) {
        for (i32 c = 1; c <= k; c++) {
          dp[i - 1][c] += dp[i - 1][c - 1];
        }
        if (i == n) break;
        for (i32 c = 0; c <= k; c++) {
          dp[i][c] += (dp[i - 1][c] - (c - i - 1 >= 0 ? dp[i - 1][c - i - 1] : 0));
          dp[i][c] %= MOD;
        }
      }
      i64 Ans = dp[n - 1][k] - (k - 1 >= 0 ? dp[n - 1][k - 1] : 0);
      Ans %= MOD;
      Ans += MOD;
      Ans %= MOD;
      return Ans;
    }
};
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