<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>

#define ll long long
ll mod = 1e9 + 7;
class Solution {
public:
    ll helper(ll index, ll cnt, vector<vector<ll>> &dp){
        if(index < 0 || (index * (index + 1)) < cnt)
            return cnt == 0;
        if(dp[index][cnt] != -1)
            return dp[index][cnt];
        dp[index][cnt] = helper(index - 1, cnt, dp);
        if(cnt > 0)
            dp[index][cnt] = (dp[index][cnt] + helper(index, cnt - 1, dp)) % mod;
        if(cnt > index)
            dp[index][cnt] = (dp[index][cnt] - helper(index - 1, cnt - index - 1, dp) + mod) % mod;
        return dp[index][cnt];
    }
    int kInversePairs(int n, int k) {
        vector<vector<ll>>dp(n, vector<ll>(k + 1, -1));
        return (int)(helper(n - 1, k, dp) % mod);
    }
};
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