<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:
        dp = {}
        def solve(n,last,count):
            if n == 0: return 1
            if (n,last,count) in dp: return dp[(n,last,count)]
            ans = 0
            for i in range(6):
                if last == i:
                    if count == rollMax[i]: continue
                    ans += solve(n-1,last,count + 1)
                else:
                    ans += solve(n-1,i,1)
            dp[(n,last,count)] = ans
            return ans
        
        return solve(n,None,0) % 1000000007
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