Problem:
Score : 100 points

Problem Statement

Takahashi is a user of a site that hosts programming contests.

When a user competes in a contest, the rating of the user (not necessarily an integer) changes according to the performance of the user, as follows:

Let the current rating of the user be a.

Suppose that the performance of the user in the contest is b.

Then, the new rating of the user will be the avarage of a and b.

For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000.

Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest.

Find the performance required to achieve it.

Constraints

0 \leq R, G \leq 4500

All input values are integers.

Input

Input is given from Standard Input in the following format:

R
G

Output

Print the performance required to achieve the objective.

Sample Input 1

2002
2017

Sample Output 1

2032

Takahashi's current rating is 2002.

If his performance in the contest is 2032, his rating will be the average of 2002 and 2032, which is equal to the desired rating, 2017.

Sample Input 2

4500
0

Sample Output 2

-4500

Although the current and desired ratings are between 0 and 4500, the performance of a user can be below 0.

Solution in Java (OpenJDK 1.8.0) (judge verdict: Accepted)



import java.io.IOException;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) throws IOException {

		Scanner scan = new Scanner(System.in);

		int R_now = scan.nextInt();
		int G_goal = scan.nextInt();

		System.out.println(2 * G_goal - R_now);

	}

	//***********************以下共通関数************************************************
	/**
	 * 文字列を反転して返す
	 * @param pStr 文字列
	 * @return 反転した文字列
	 */
	public static String myReverse(String pStr) {
		if (pStr.length() < 1) {
			return "";
		}

		StringBuffer sb = new StringBuffer(pStr);
		return sb.reverse().toString();
	}

	/**
	 * 余りを求める。
	 * @param a
	 * @param b
	 * @return 余り。余りがマイナスの時はプラスで返す。
	 */
	public static long myMod(long a, long b) {
		long ret = 0;

		ret = a % b;

		if (ret < 0) {
			ret += b;
		}

		return ret;
	}

	/**
	 * 乗数を求める ※mod対応
	 * @param kisu 基数
	 * @param josu 乗数
	 * @param primeNumMod_0 法 mod値 使わない場合は「0」
	 * @return
	 */
	public static long myPowWithMod(long kisu, long josu, long primeNumMod_0) {

		long ans = 1;

		//乗数が0の場合→1
		if (josu == 0)
			return 1;

		if (primeNumMod_0 == 0) {
			//modを使わないとき

			while (josu > 0) {

				if (josu % 2 == 0) {
					//累乗が偶数の時

					kisu *= kisu;
					josu = josu / 2;
				} else {
					//累乗が奇数の時
					ans *= kisu;
					josu--;
				}
			}

			return ans;

		} else {
			//modを使うとき

			//初めに桁を落とす
			kisu = myMod(kisu, primeNumMod_0);

			while (josu > 0) {

				if (josu % 2 == 0) {
					//累乗が偶数の時

					kisu *= myMod(kisu, primeNumMod_0);
					josu = josu / 2;

				} else {
					//累乗が奇数の時
					ans *= myMod(kisu, primeNumMod_0);
					josu--;

				}
			}

			return myMod(ans, primeNumMod_0);

		}

	}

}