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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
// (setq-default c-basic-offset 2)

int main() {
  int x, y;
  cin >> x >> y;

  // c has 2 legs
  // t has 4 legs

  // x animals
  // y legs

  //  c +  t = x
  // 2c + 4t = y

  // x = c + t
  // y = 2 * (c + 2t)  --> y is an even number

  //  c +  t = x
  // 2c + 4t = y

  // 4c + 4t = 4x    <-->    4t = 4x - 4c
  // 2c + 4t = y

  // 4c + 4t = 4x      <-->    4t = 4x - 4c
  // 2c + 4x - 4c = y  <-->    -2c + 4x = y
  //                   <-->     2c = 4x - y   <--> c = 2x - y/2 = (4x - y) / 2

  // c = (4x - y) / 2
  // t = x - c
  //   = x - 2x + y/2
  //   = y/2 - x
  // t = (y - 2x) / 2

  // c = (4x - y) / 2   --> 4x >= y
  // t = (y - 2x) / 2   --> y >= 2x    --> 2x <= y <= 4x

  if (y % 2 != 0) {
    cout << "No" << endl;
    return 0;
  }
  if (2 * x <= y and y <= 4 * x) {
    cout << "Yes" << endl;
    return 0;
  } else {
    cout << "No" << endl;
    return 0;
  }

  return 0;
}