Adding fingers P55691

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:

1. Add mod 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:

   ⎧ ⎪ ⎨ ⎪ ⎩

   (a,b)(c,d) → (a,b)(c+a,d) if  a,c ≠ 0          (a,b)(c,d) → (a,b)(c,d+a) if  a,d ≠ 0          (a,b)(c,d) → (a,b)(c+b,d) if  b,c ≠ 0          (a,b)(c,d) → (a,b)(c,d+b) if  b,d ≠ 0

2. “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 mod 5:

   ⎧ ⎨ ⎩

   (a,b)(c,d) → (a+b,0)(c,d) if  a,b ≠ 0      (a,b)(c,d) → (0,a+b)(c,d) if  a,b ≠ 0

3. “Redistribute” the fingers in X’s hands, if one of them is empty:

   ⎧ ⎨ ⎩

   (a,0)(c,d) → (x,y)(c,d) if  x+y=a  and  0 < x,y < a          (0,b)(c,d) → (x,y)(c,d) if  x+y=b  and  0 < x,y < b

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.