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: Wrong Answer)

//Hiia
#include <stdio.h>
#include <stdlib.h>//rannsuu
#include <time.h>
//srand((unsigned int) time(0));
#include <math.h>//koudona keisann
#include <string.h>//hairetsu
//#define N

//scanf("\n", );
//printf("\n", );

//long n,s;
//scanf("%ld %ld", &n, &s);

//for(i = 0; i < N; i++) {

//long long a [100010];
//for(int i=0;i<n;i++) scanf("%lld",&a[i]);


int main(){
  int X, Y, n;
  scanf("%d%d", &X,&Y);
  if(4*X-Y % 2 != 1){
    printf("No\n");
  }
  if(4*X-Y % 2 != 0){
    n=(4*X-Y)/2;
    if(0<=n || n<=X){
      printf("Yes\n");
    }
    if(0>n || n>X){
      printf("No\n");
    }
  }



  return 0;
}