It is possible to show that the square root of two can be expressed as an infinite continued fraction.

$\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots }}}$

By expanding this for the first four iterations, we get:

$1+\frac{1}{2}=\frac{3}{2}=1.5$
$1+\frac{1}{2+\frac{1}{2}}=\frac{7}{5}=1.4$
$1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}=\frac{17}{12}=1.41666\dots$
$1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}=\frac{41}{29}=1.41379\dots$

The next three expansions are $\frac{99}{70}$, $\frac{239}{169}$, and $\frac{577}{408}$, but the eighth expansion, $\frac{1393}{985}$, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than the denominator?