With $2018$ points, a network composed of hexagons is formed as a sample the figure: [asy][asy] unitsize(1 cm); int i; path hex = dir(30)--(0,1)--dir(150)--dir(210)--(0,-1)--dir(330)--cycle; draw(hex); draw(shift((sqrt(3),0))*(hex)); draw(shift((2*sqrt(3),0))*(hex)); draw(shift((4*sqrt(3),0))*(hex)); draw(shift((5*sqrt(3),0))*(hex)); dot((3*sqrt(3) - 0.3,0)); dot((3*sqrt(3),0)); dot((3*sqrt(3) + 0.3,0)); dot(dir(150)); dot(dir(210)); for (i = 0; i <= 5; ++i) { if (i != 3) { dot((0,1) + i*(sqrt(3),0)); dot(dir(30) + i*(sqrt(3),0)); dot(dir(330) + i*(sqrt(3),0)); dot((0,-1) + i*(sqrt(3),0)); } } dot(dir(150) + 4*(sqrt(3),0)); dot(dir(210) + 4*(sqrt(3),0)); [/asy][/asy] A bee and a fly play the following game: initially the bee chooses one of the $2018$ dots and paints it red, then the fly chooses one of the $2017$ unpainted dots and paint it blue. Then the bee chooses an unpainted point and paints it red and then the fly chooses an unpainted one and paints it blue and so they alternate. If at the end of the game there is an equilateral triangle with red vertices, the bee wins, otherwise Otherwise the fly wins. Determine which of the two insects has a winning strategy.