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 11.0.6) (judge verdict: Accepted)

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

public class Main {
	public static void main(String[] args) throws IOException {
//		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
//		String W	 = in.readLine();
//		int num = Integer.parseInt(W.split(" ")[0]);

		Scanner sc = new Scanner(System.in);
		int X = sc.nextInt();
		int Y = sc.nextInt();
		boolean ans = false;

		int tsuru = X;
		int kame = 0;
		for(;tsuru >=0;tsuru--) {
			kame=X-tsuru;
			if(tsuru*2 + kame*4 ==Y) {
			ans = true;
			}
		}



		if(ans) {
		System.out.println("Yes");
		}else {
			System.out.println("No");
		}

	}

	//	//二分探索
	//k <= num となる最小の配列要素kのインデックスを返す
	static private int binarySearch(long num, long[] orderedArray){
		int lowerBorder = -1;
		int upperBorder = orderedArray.length;
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedArray[mid]<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//二分探索
	//k <= num となる最小のList要素kのインデックスを返す
	static private int binarySearch(long num, ArrayList<Long> orderedList){
		int lowerBorder = -1;
		int upperBorder = orderedList.size();
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedList.get(mid)<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//aとbの最小公倍数を求める
	public static int gcd(int a, int b) {
        return b == 0 ? a: gcd(b, a % b);
	}
	public static long gcd(long a, long b) {
        return b == 0 ? a: gcd(b, a % b);
	}



}