\(\sqrt{-1}\) isn't imaginary!

If \(x_1,x_2,\ldots, x_i\) are the distinct solutions to the following congruence (with \(0\leq x_k<13\) for each \(k\), \(k \in \{1,2,\ldots ,i\}\)), \[\large x^2 \equiv -1 \pmod{13}\]

then enter the remainder when \((x_1+x_2+\cdots+x_i)\) is divided by 13.

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