<p>A die simulator generates a random number from <code>1</code> to <code>6</code> for each roll. You introduced a constraint to the generator such that it cannot roll the number <code>i</code> more than <code>rollMax[i]</code> (<strong>1-indexed</strong>) consecutive times.</p>

<p>Given an array of integers <code>rollMax</code> and an integer <code>n</code>, return <em>the number of distinct sequences that can be obtained with exact </em><code>n</code><em> rolls</em>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Two sequences are considered different if at least one element differs from each other.</p>

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

<pre>
<strong>Input:</strong> n = 2, rollMax = [1,1,2,2,2,3]
<strong>Output:</strong> 34
<strong>Explanation:</strong> There will be 2 rolls of die, if there are no constraints on the die, there are 6 * 6 = 36 possible combinations. In this case, looking at rollMax array, the numbers 1 and 2 appear at most once consecutively, therefore sequences (1,1) and (2,2) cannot occur, so the final answer is 36-2 = 34.
</pre>

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

<pre>
<strong>Input:</strong> n = 2, rollMax = [1,1,1,1,1,1]
<strong>Output:</strong> 30
</pre>

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

<pre>
<strong>Input:</strong> n = 3, rollMax = [1,1,1,2,2,3]
<strong>Output:</strong> 181
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 5000</code></li>
	<li><code>rollMax.length == 6</code></li>
	<li><code>1 &lt;= rollMax[i] &lt;= 15</code></li>
</ul>

class Solution:
    def dieSimulator(self, n: int, rollMax: List[int]) -> int:
        cache = {}
        MOD = 1000000007
        def helper(k, prev):
            if (k, prev) in cache:
                return cache[(k, prev)]
            if k == 0:
                cache[(k, prev)] = 1
                return 1
            
            res = 0
            for val in range(6):
                if val == prev:
                    continue
                for freq in range(1, min(rollMax[val], k) + 1):
                    res += helper(k-freq, val)
            
            res %= MOD
            cache[(k, prev)] = res
            return res
        
        return helper(n, -1)
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