<p>You are given three integers <code>n</code>, <code>m</code> and <code>k</code>. Consider the following algorithm to find the maximum element of an array of positive integers:</p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/02/e.png" style="width: 424px; height: 372px;" />
<p>You should build the array arr which has the following properties:</p>

<ul>
	<li><code>arr</code> has exactly <code>n</code> integers.</li>
	<li><code>1 &lt;= arr[i] &lt;= m</code> where <code>(0 &lt;= i &lt; n)</code>.</li>
	<li>After applying the mentioned algorithm to <code>arr</code>, the value <code>search_cost</code> is equal to <code>k</code>.</li>
</ul>

<p>Return <em>the number of ways</em> to build the array <code>arr</code> under the mentioned conditions. As the answer may grow large, the answer <strong>must be</strong> computed modulo <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 2, m = 3, k = 1
<strong>Output:</strong> 6
<strong>Explanation:</strong> The possible arrays are [1, 1], [2, 1], [2, 2], [3, 1], [3, 2] [3, 3]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 5, m = 2, k = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no possible arrays that satisfy the mentioned conditions.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 9, m = 1, k = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only possible array is [1, 1, 1, 1, 1, 1, 1, 1, 1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>1 &lt;= m &lt;= 100</code></li>
	<li><code>0 &lt;= k &lt;= n</code></li>
</ul>

class Solution {
public:
    int numOfArrays(int n, int m, int k)
    {
        int kk=k;
        long long dp[n+1][m+1][kk+2];
        int i,j,p;
        long long brr[m+1][kk+2];
        int mod=1e9+7;
        for(i=0;i<=n;i++)
        {
            for(j=0;j<=m;j++)
            {
                for(p=0;p<=kk;p++)
                dp[i][j][p]=0;
                for(p=0;p<=kk;p++)
                brr[j][p]=0;
            }
        }
        for(i=1;i<=n;i++)
        {
            if(i==1)
            {
                for(j=1;j<=m;j++)
                dp[i][j][0]=1;
            }
            else
            {
                for(j=1;j<=m;j++)
                {
                    for(p=0;p<=kk;p++)
                    dp[i][j][p]+=(dp[i-1][j][p]*j)%mod;
                }
                for(j=1;j<=m;j++)
                {
                    for(p=0;p<=kk;p++)
                    dp[i][j][p+1]=(dp[i][j][p+1]+brr[j-1][p])%mod;
                }
            }
            for(p=0;p<=kk;p++)
                {
                    long long sum=0;
                    for(j=1;j<=m;j++)
                    {
                        sum=sum+dp[i][j][p];
                        brr[j][p]=sum;
                    }
                }
        }
        long long ans=0;
        for(j=1;j<=m;j++)
        ans=(ans+dp[n][j][kk-1])%mod;
        return ans;
    }
};
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