Lazy Snakes and Ladders (2) P65875

The author of the previous problem is half a psycho, half a freak. He is well know for his lengthy and weird statements, although a few of his problems are not that hard. Consider for instance the following adaptation of an old problem of the same author.

The rules of Snakes and Ladders (see the board below) are simple:

• If the landing cell (after potential bouncing) is the bottom of a ladder, the counter is moved to its top, which will be a higher-numbered cell (e.g., rolling a 1 when on cell 1 takes the counter to 38). Nothing happens when the counter directly lands on a top.
• If the landing cell (after potential bouncing) is the head of a snake, the counter is moved to its tail, which will be a lower-numbered one (e.g., rolling a 3 when on cell 98 takes the counter to 80). Nothing happens when the counter directly lands on a tail.
• If the rolled number was six, the player keeps the turn; otherwise, it passes to the other player (irrespective of whether bouncing, snakes, or ladders were involved).

You must simulate several of these games using pseudo-random numbers. In particular, include the <random> library, and declare a global

mt19937 rng;

variable. Every game will be defined by a seed s. Just do

rng.seed(s);

to reset rng before every game. Afterwards, every time that you need the result d of the next rolling of the die, use this code:

unsigned int r = rng();
int d = r%6 + 1;

For instance, with the initial seed 42, we get these values for r: 1608637542, 3421126067, 4083286876, 787846414, …Therefore, the values for d are 1, 6, 5, 5, …In this game, red goes to 2 (and then to 38), blue goes to 7 (and then to 14), blue moves again (he got a 6) and goes to 19, red goes to 43, …, and eventually blue wins.

Input

Input consists of several games, each one defined by an integer seed s between 1 and 10⁹.

Output

For each game, print “RED” or “BLUE” depending on the winner.