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Vapid01 instance is not accepted as a key object #137

@mthu

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@mthu

Since Vapid01 is a superclass of Vapid02 (and thus Vapid) in py_vapid library, following condition does not match the Vapid01 instance and the ongoing code treats the object as a string (and fails parsing it).

if isinstance(vapid_private_key, Vapid):

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