<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 kInversePairs(long long n, long long k) {
        long long N = 1e9+7;
        long long dp[n][k+1];
        for(long long i=0;i<n;i++){
            for(long long j=0;j<=k;j++){
                dp[i][j]=0;
            }
        }
        dp[0][0]=1;
        vector<long long> pref(k+2,1);pref[0]=0;
        long long prev_store=1;
        for(int i=1;i<n;i++){
            for(int j=0;j<=k;j++){
                long long ans = 0;
                ans=pref[j+1]-pref[max(0,j-i)];
                ans=(ans+N)%N;
                dp[i][j]=ans;
            }

            for(int j=1;j<k+2;j++){
                pref[j]=pref[j-1]+dp[i][j-1];
            }
        }
        return dp[n-1][k];
    }
};
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