Abbreviating the Product of a Range LeetCode solution in Python

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class Solution:
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    def abbreviateProduct(self, left: int, right: int) -> str:
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        c2 = c5 = 0
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        top12 = tail5 = 1
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        for i in range(left, right + 1):
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            # count and remove all 2 and 5
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            while i % 2 == 0:
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                i //= 2
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                c2 += 1
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            while i % 5 == 0:
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                i //= 5
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                c5 += 1
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            # track top 12 and last 5
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            top12 = int(str(top12 * i)[:12])
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            tail5 = tail5 * i % 100000
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        # multiply the remained 2 or 5
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        if c2 > c5:
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            for _ in range(c2 - c5):
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                top12 = int(str(top12 * 2)[:12])
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                tail5 = tail5 * 2 % 100000
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        elif c2 < c5:
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            for _ in range(c5 - c2):
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                top12 = int(str(top12 * 5)[:12])
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                tail5 = tail5 * 5 % 100000
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        zero = min(c2, c5)
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        # as is included in top 12, it's easy to tell when d<=10
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        if len(str(top12)) <= 10:
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            return str(top12) + "e" + str(zero)
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        return (
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            str(top12)[:5]
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            + "." * 3
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            + "0" * (5 - len(str(tail5)))
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            + str(tail5)
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            + "e"
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            + str(zero)
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        )