<p>Your task is to calculate <code>a<sup>b</sup></code> mod <code>1337</code> where <code>a</code> is a positive integer and <code>b</code> is an extremely large positive integer given in the form of an array.</p>

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

<pre>
<strong>Input:</strong> a = 2, b = [3]
<strong>Output:</strong> 8
</pre>

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

<pre>
<strong>Input:</strong> a = 2, b = [1,0]
<strong>Output:</strong> 1024
</pre>

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

<pre>
<strong>Input:</strong> a = 1, b = [4,3,3,8,5,2]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= a &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>1 &lt;= b.length &lt;= 2000</code></li>
	<li><code>0 &lt;= b[i] &lt;= 9</code></li>
	<li><code>b</code> does not contain leading zeros.</li>
</ul>

class Solution:
    def superPow(self, a: int, b: List[int]) -> int:
        result = 1
        for num in b:
            result = (pow(result, 10, 1337) * pow(a, num, 1337)) % 1337
        return result

# [1, 10, 100]

# (\U0001d44e\U0001d44fmod\U0001d45a)=((\U0001d44emod\U0001d45a)(\U0001d44fmod\U0001d45a))mod\U0001d45a


# 2^(433852)

# 2 ^ 2 
# 2 ^ 50 
# 2 ^ 800
# 2 ^ 3000
# 2 ^ 30000 = 2 ^ 3 * 2 ^ 10000 = 2 ^ (3 * 10000)
# 2 ^ 400000

16600