Problem:
Score : 200 points

Problem Statement

There are some animals in a garden. Each of them is a crane with two legs or a turtle with four legs.

Takahashi says: "there are X animals in total in the garden, and they have Y legs in total." Determine whether there is a combination of numbers of cranes and turtles in which this statement is correct.

Constraints

1 \leq X \leq 100

1 \leq Y \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

If there is a combination of numbers of cranes and turtles in which the statement is correct, print Yes; otherwise, print No.

Sample Input 1

3 8

Sample Output 1

Yes

The statement "there are 3 animals in total in the garden, and they have 8 legs in total" is correct if there are two cranes and one turtle. Thus, there is a combination of numbers of cranes and turtles in which the statement is correct.

Sample Input 2

2 100

Sample Output 2

No

There is no combination of numbers of cranes and turtles in which this statement is correct.

Sample Input 3

1 2

Sample Output 3

Yes

We also consider the case in which there are only cranes or only turtles.

Solution in Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int X = scanner.nextInt();
        int Y = scanner.nextInt();

        int[][] A = {{1, 1, X}, {4, 2, Y}};

        if (Y / X == 2) {
            System.out.println("Yes");
        } else if (Y / X == 4) {
            System.out.println("No");
        } else {
            for (int k = 0; k < 2; k++) { // ステップ１～ステップ４
                double p = A[k][k];// ピボット係数
                int i;
                for (i = k; i < 2; i++)
                    A[k][i] /= p;  // ピボット係数を１にするためピボット行を割り算

                for (i = 0; i < 2; i++) {// ピボット列の掃き出し
                    if (i != k) {
                        double d = A[i][k];
                        for (int j = k; j < 2; j++) {
                            A[i][j] -= d * A[k][j];
                            if (A[i][j] != 1 | A[i][j] != 0) {
                                System.out.println("No");
                                System.exit(0);
                            }
                        }
                    }
                }
            }
            System.out.println("Yes");
        }
    }
}