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>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds; 
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
 
    int x,y;
    cin>>x>>y;
    int ans;
    for(int i=1;i<=x;i++)
    {
        ans = i*2;
        ans+=abs((x-i))*4;
        if(ans == y)
        {
            cout<<"Yes";
            return 0;
        }
    }
    if((x*2) == y || (x*4) == y)
        cout<<"Yes";
    else
        cout<<"No";
 
    return 0;
}