from math import log10
 
n, d = 3, 2
tot = 0
for i in xrange(1000):
    if int(log10(n)) > int(log10(d)):
        tot += 1
    n,d = n+2*d, n+d
 
print tot

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