Adding fingers

Consider a game for two players playing alternatively. Both players show a certain number of fingers in each hand. Let X be the player that moves next, and let Y be the other player. Let $a$ and $b$ be the number of fingers shown by X, and let $c$ and $d$ be the number of fingers shown by Y. In each turn, these are the allowed moves:

Add $\bmod$ 5 as many fingers as X has in a non-empty hand (a hand showing at least one finger) to one of Y’s non-empty hands. That is: $\left\{\begin{array}{cc} (a,b)(c,d) \to (a,b)(c+a,d) & \text{if } a,c \neq 0 \\ (a,b)(c,d) \to (a,b)(c,d+a) & \text{if } a,d \neq 0 \\ (a,b)(c,d) \to (a,b)(c+b,d) & \text{if } b,c \neq 0 \\ (a,b)(c,d) \to (a,b)(c,d+b) & \text{if } b,d \neq 0 \end{array}\right.$
“Move” the fingers in one of X’s hands to the other hand, provided that none of them are empty. Again, the operations are made $\bmod$ 5: $\left\{\begin{array}{lr} (a,b)(c,d) \to (a+b,0)(c,d) & \text{if } a,b \neq 0 \\ (a,b)(c,d) \to (0,a+b)(c,d) & \text{if } a,b \neq 0 \end{array}\right.$
“Redistribute” the fingers in X’s hands, if one of them is empty: $\left\{\begin{array}{lr} (a,0)(c,d) \to (x,y)(c,d) & \text{if } x+y=a \text{ and } 0 < x,y < a \\ (0,b)(c,d) \to (x,y)(c,d) & \text{if } x+y=b \text{ and } 0 < x,y < b \end{array}\right.$

Both players play perfectly. The first player to get to $(0, 0)$ loses the game. A game that never ends is considered to be a draw.

Input

Input consists of several cases, each one with $a$ , $b$ , $c$ and $d$ , all between 0 and 4. Assume $a + b > 0$ and $c + d > 0$ .

Output

For every case, tell if X will win, if X will lose, or if the game is a draw.

Problem information

Author: Marc Felipe

Generation: 2026-01-25T11:19:59.298Z