Integer Game
Contest: Codeforces Global Round 9
Difficulty rating: 2600
Tags: constructive algorithms, games, interactive, math

This is an interactive problem.

Anton and Harris are playing a game to decide which of them is the king of problemsetting.

There are three piles of stones, initially containing $$$a$$$, $$$b$$$, and $$$c$$$ stones, where $$$a$$$, $$$b$$$, and $$$c$$$ are distinct positive integers. On each turn of the game, the following sequence of events takes place:

- The first player chooses a positive integer $$$y$$$ and provides it to the second player.
- The second player adds $$$y$$$ stones to one of the piles, with the condition that he cannot choose the same pile in two consecutive turns.

The second player loses if, at any point, two of the piles contain the same number of stones. The first player loses if $$$1000$$$ turns have passed without the second player losing.

Feeling confident in his skills, Anton decided to let Harris choose whether he wants to go first or second. Help Harris defeat Anton and become the king of problemsetting!

Input format:
The first line of input contains three distinct positive integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \le a, b, c \le 10^9$$$)  — the initial number of stones in piles $$$1$$$, $$$2$$$, and $$$3$$$ respectively.

Interaction format:
The interaction begins by reading the integers $$$a$$$, $$$b$$$ and $$$c$$$.

After reading the integers, print a single line containing either "First" or "Second", denoting who you want to play as (as first or second correspondently).

On each turn, the first player (either you or the judge) must print a positive integer $$$y$$$ ($$$1 \le y \le 10^{12}$$$).

Then, the second player must print $$$1$$$, $$$2$$$, or $$$3$$$, indicating which pile should have $$$y$$$ stones added to it. From the second turn onwards, the pile that the second player chooses must be different from the pile that they chose on the previous turn.

If you are playing as Second and complete $$$1000$$$ turns without losing, or if you are playing as First and the judge has determined that it cannot make a move without losing, the interactor will print $$$0$$$ and will finish interaction. This means that your program is correct for this test case, and you should exit immediately.

If you are playing as First and complete $$$1000$$$ turns without winning, or if you are playing as Second and print a move that makes two piles have the same number of stones, or if you output an invalid move as either player, the interactor will print $$$-1$$$ and will finish interaction. You will receive a Wrong Answer verdict. Make sure to exit immediately to avoid getting other verdicts.

After printing something do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:

- fflush(stdout) or cout.flush() in C++;
- System.out.flush() in Java;
- flush(output) in Pascal;
- stdout.flush() in Python;
- see documentation for other languages.

In this problem, hacks are disabled.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the sample input, the piles initially have $$$5$$$, $$$2$$$, and $$$6$$$ stones. Harris decides to go first and provides the number $$$2$$$ to Anton. Anton adds $$$2$$$ stones to the third pile, which results in $$$5$$$, $$$2$$$, and $$$8$$$.

In the next turn, Harris chooses $$$3$$$. Note that Anton cannot add the stones to the third pile since he chose the third pile in the previous turn. Anton realizes that he has no valid moves left and reluctantly recognizes Harris as the king.

Example input:
5 2 6


3

0

Example output:
First
2

3

Editorial:
Let's prove that the first player can always win in at most three turns. Assume that the initial numbers of stones are $$$p < q < r$$$. On the first turn, choose $$$y = 2r - p - q$$$. We then have three cases:
Case 1.  $$$y$$$ is added to $$$p$$$.
The piles now have $$$2r - q$$$, $$$q$$$ and $$$r$$$ stones. Now choose $$$y = r - q$$$. Since the pile with $$$2r - q$$$ stones cannot be chosen on this turn, this results in a win.
Case 2.  $$$y$$$ is added to $$$q$$$.
The piles now have $$$p$$$, $$$2r - p$$$ and $$$r$$$ stones. Choose $$$y = r - p$$$. Similarly to the previous case this results in a win since the pile with $$$2r - p$$$ stones cannot be chosen on this turn.
Case 3.  $$$y$$$ is added to $$$r$$$.
The piles have $$$p < q < 3r - p - q$$$ stones. We now have a situation similar to the initial one, with the difference that the pile with the largest number of stones cannot be chosen on the next turn. Thus we may repeat the strategy and obtain a guaranteed win this time.