<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:
    long long mod = 1e9 + 7;
    long long findMultipleUptoNextI(int size, long long num) {
        long long result = 1;
        while(size--) {
            result *= num%mod;
            result %= mod;
        }
        return result%mod;
    }
    long long solve(int& n, int& m, int& k, int current_i, int current_k, long long current_m, vector<vector<vector<long long>>>& dp) {
        if(current_i >= n) {
            return 0;
        }
        if(current_k == k) {
            long long temp = findMultipleUptoNextI(n-current_i-1, current_m);
            temp %= mod;
            return temp;
        }
        if(dp[current_i][current_m][current_k]!=-1) {
            return dp[current_i][current_m][current_k]%mod;
        }
        long long ans = 0;
        for(int index = current_i+1; index < n; index++) {
            long long temp = findMultipleUptoNextI(index - current_i-1, current_m)%mod;
            temp = temp%mod;
            long long temp2 = temp;
            for(long long num = current_m+1; num <= m; num++) {
                temp *= solve(n, m, k, index, current_k+1, num, dp)%mod;
                temp %= mod;
                ans += temp%mod;
                ans %= mod;
                temp = temp2%mod;
            }
        }
        dp[current_i][current_m][current_k] = ans%mod;
        return ans%mod;
    }
    int numOfArrays(int n, int m, int k) {
        long long answer = 0;
        vector<vector<vector<long long>>> dp(n+1, vector<vector<long long>>(m+1, vector<long long>(k+1,-1)));
        for(int i = 1; i <= m; i++) {
            answer += solve(n, m, k, 0, 1, i, dp)%mod;
            answer %= mod;
        }  
        return answer%mod;
    }
};
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