Problem: 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

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

Interaction

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 ≤ y ≤ 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.

Example

Input


5 2 6


3

0

Output



First
2

3

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.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class f {
	static final FS sc = new FS();
	static final PrintWriter pw = new PrintWriter(System.out);
	
	static final int n = 3, numTurns = 1_000, turnsToEnd = 10;
	
	static long[] orig, curr;
	static int prevI, turn;
	static long lastY;
	public static void main(String[] args) {
		orig = new long[n];
		curr = new long[n];
		for(int i = 0; i < n; ++i) curr[i] = orig[i] = sc.nextLong();
		
		pw.println("First");
		pw.flush();
		prevI = -1;
		lastY = -1;
		for(turn = 0; turn < numTurns; ++turn) {
//			System.out.println(Arrays.toString(curr));
			if(first(turn)) {
				pw.println(lastY);
				pw.flush();
				
				int pileIdx = sc.nextInt() - 1;
				
				// Correct!
				if(pileIdx == -1) return;
				// Wrong Answer
				else if(pileIdx == -2) {
					WrongAnswer();
					return;
				}
				
				curr[pileIdx] += lastY;
				prevI = pileIdx;
			}
			else {
				// hopefully doesn't happen
				WrongAnswer();
				return;
			}
		}
		
		WrongAnswer();
	}
	static boolean first(int tempTurn) {
		if(tempTurn == turnsToEnd) return false;
		
		long[] sort = new long[n];
		for(int i = 0; i < n; ++i) sort[i] = curr[i];
		Arrays.sort(sort);
		
		ArrayList<Long> al = new ArrayList<>();
		long d1 = sort[1] - sort[0], d2 = sort[2] - sort[1];
		// kill bottom on middle
		add(al, d1);
		// kill bottom on top
		add(al, d1 + d2);
		// kill middle on top
		add(al, d2);
		
		// set up triplets:
		
		// get bottom to first
		add(al, d1 - d2);
		// get bottom to middle
		add(al, d1 + d2 / 2);
		// get bottom to last
		add(al, d1 + d2 + d2);
		
		// get middle to actual middle
		add(al, (d2 - d1) / 2);
		// get middle to last; (we got this one already)
//		add(al, d1 + d2 + d2);
		
		// get last to actual last
		add(al, d2 - d1);
		
		for(long y : al) {
			if(!second(y, tempTurn)) {
				if(tempTurn == turn) lastY = y;
				return true;
			}
		}
		
		return false;
	}
	static boolean second(long y, int tempTurn) {
		HashSet<Long> got = new HashSet<>();
		for(int i = 0; i < n; ++i) got.add(curr[i]);
		
		for(int i = 0; i < n; ++i) {
			if(i != prevI && !got.contains(curr[i] + y)) {
				boolean won = false;
				
				int tempPrevI = prevI;
				// try adding here
				curr[i] += y;
				prevI = i;
				if(!first(tempTurn + 1)) won = true;
				curr[i] -= y;
				prevI = tempPrevI;
				
				if(won) return true;
			}
		}
		
		return false;
	}
	static void add(ArrayList<Long> al, long y) {
		if(y > 0) al.add(y);
	}
	static void WrongAnswer() {
		pw.println(orig[0] + "-" + orig[1] + "-" + orig[2] + ":WrongAnswer");
		pw.flush();
	}
	static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while(!st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}
/*
5 2 6

*/