<p>Given an integer <code>n</code>, your task is to count how many strings of length <code>n</code> can be formed under the following rules:</p>

<ul>
	<li>Each character is a lower case vowel&nbsp;(<code>&#39;a&#39;</code>, <code>&#39;e&#39;</code>, <code>&#39;i&#39;</code>, <code>&#39;o&#39;</code>, <code>&#39;u&#39;</code>)</li>
	<li>Each vowel&nbsp;<code>&#39;a&#39;</code> may only be followed by an <code>&#39;e&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;e&#39;</code> may only be followed by an <code>&#39;a&#39;</code>&nbsp;or an <code>&#39;i&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;i&#39;</code> <strong>may not</strong> be followed by another <code>&#39;i&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;o&#39;</code> may only be followed by an <code>&#39;i&#39;</code> or a&nbsp;<code>&#39;u&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;u&#39;</code> may only be followed by an <code>&#39;a&#39;</code>.</li>
</ul>

<p>Since the answer&nbsp;may be too large,&nbsp;return it modulo&nbsp;<code>10^9 + 7</code>.</p>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 5
<strong>Explanation:</strong> All possible strings are: &quot;a&quot;, &quot;e&quot;, &quot;i&quot; , &quot;o&quot; and &quot;u&quot;.
</pre>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 10
<strong>Explanation:</strong> All possible strings are: &quot;ae&quot;, &quot;ea&quot;, &quot;ei&quot;, &quot;ia&quot;, &quot;ie&quot;, &quot;io&quot;, &quot;iu&quot;, &quot;oi&quot;, &quot;ou&quot; and &quot;ua&quot;.
</pre>

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

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 68</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 2 * 10^4</code></li>
</ul>

class Solution:
    def countVowelPermutation(self, n: int) -> int:
        relations = {
            '.': ['a', 'e', 'i', 'o', 'u'],
            'a': ['e'],
            'e': ['a', 'i'],
            'i': ['a', 'e', 'o', 'u'],
            'o': ['i', 'u'],
            'u': ['a']
        }

        @lru_cache(None)
        def dfs(i, prev):
            if i >= n:
                return 1
            
            ans = 0
            for nxt in relations[prev]:
                ans += dfs(i + 1, nxt)
            return ans 
        
        return dfs(0, '.') % (10 ** 9 + 7)
193597