Given a sequence of numbers a1,a2,...,a15, one can always construct a new sequence b1,b2,...,b15, where bi is equal to the number of terms in the sequence {ak}k=115 less than ai (i=1,2,...,15). Is there a sequence {ak}k=115 for which the sequence {bk}k=115 is 1,0,3,6,9,4,7,2,5,8,8,5,10,13,13?