<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:
        
        MOD = 1337
        def power_mod(x,n):
            result =1
            x %=MOD
            while n>0:
                if n&1:
                    result = (result * x)%MOD
                x = (x*x)%MOD
                n>>=1

            return result
        
        if not b:
            return 1
        
        last_digit = b.pop()
        return (power_mod( self.superPow(a,b), 10) * power_mod(a,last_digit)) %MOD

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