Consider the set \(S=\{1,2,3,\cdots,101\}\), and let \(S_k\) be the sum of the product of every \(k\) element subset of \(S\) (also called the \(k\)th symmetric sum of \(S\)). Find the remainder when \(\sum_{k=1}^{101} S_k\) is divided by \(101\).

This problem is posed by Daniel C.

**Details and assumptions**

You may use the fact that 101 is prime.

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