Index of a Farey sequence

Let $F_{idx}(n,k)$ denote the $k$-th term in the Farey sequence of order $n$. For example, the Farey sequence of order 5 is {0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1}, and the 7th term is 3/5, so $F_{idx}(5,7)=3/5$. You are also given $F_{idx}(12,31)=7/11$, $F_{idx}(123,3141)=82/121$.

Find $F_{idx}(1234567890,314159265358979323)$. The answer should be represented as an irreducible fraction.

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Thanks to czp for the idea.