<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 {
public:
    long long dp[1010][1010];
    const int mod = 1e9+7;

    long long rec(int n, int k){
        //pruning
        if(k<0) return 0;
        if(k == 0) return 1;

        //basecase
        if(n == 0){
            if(k == 0) return 1;
            else return 0;
        }

        //cache check
        if(dp[n][k] != -1) return dp[n][k];

        //compute
        //long long ans = 0;
        // for(int i = n; i>=1; i--){
        //     ans = (mod + (ans%mod + rec(n-1, k-(n-i))%mod)%mod)%mod;
        // }

        long long ans = (mod + (rec(n-1,k)%mod + rec(n,k-1)%mod - rec(n-1,k-n)%mod)%mod)%mod;
        
        //save and return
        return dp[n][k] = ans%mod;
    }
    int kInversePairs(int n, int k) {
        memset(dp,-1,sizeof(dp));
        // for(int i = 0; i<=n; i++){
        //     for(int j = 0; j<=k; j++){
        //         cout << rec(i,j) << " ";
        //     }
        //     cout << endl;
        // }

        return rec(n,k);
    }
};
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