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

~~~
sqrt(2) = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
~~~

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

~~~
1 + 1/2                         = 3 /2  = 1.5
1 + 1/(2 + 1/2)                 = 7 /5  = 1.4
1 + 1/(2 + 1/(2 + 1/2))         = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...
~~~

The next three expansions are `99/70`, `239/169`, and `577/408`, but the eighth expansion, `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 denominator?