OME 2017 Problem 4

You are given a row made by $2018$ squares, numbered consecutively from $0$ to $2017$ . Initially, there is a coin in the square $0$ . Two players $A$ and $B$ play alternatively, starting with $A$ , on the following way: In his turn, each player can either make his coin advance $53$ squares or make the coin go back $2$ squares. On each move the coin can never go to a number less than $0$ or greater than $2017$ . The player who puts the coin on the square $2017$ wins. ¿Who is the one with the wining strategy and how should he play to win?