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)

//Author: AnandRaj doubleux
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef pair<ll,ll> pll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<pll> vpll; 

#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define test() int t;cin>>t;while(t--)
#define all(v) v.begin(),v.end()
#define prin(V) for(auto v:V) cout<<v<<" ";cout<<endl
#define take(V,f,n) for(int in=f;in<f+n;in++) cin>>V[in]
#define what(x) cerr<<#x<<" = "<<x<<endl
#define KStest() int t,t1;cin>>t;t1=t;while(t--)
#define KScout cout<<"Case #"<<t1-t<<": "

const int MOD = 1e9+7,MAX = 1e6+5;
/////////////////FastExp///////////////////
ll powN(ll a,ll p,ll mod)
{
	if(p==0) return 1;
	ll z=powN(a,p/2,mod);
	z=(z*z)%mod;
	if(p%2) z=(z*a)%mod;
	return z;
}
/////////////////FastExp///////////////////
//////////////////Sieve////////////////////
vector<bool> is_prime(MAX, true);
vector<int> MinDiv(MAX);
void Sieve()
{
	is_prime[0] = is_prime[1] = false;
	int i,j;
	for (i = 2; i*i <= MAX; i++) 
	{
    	if (is_prime[i]) 
    	{
    		MinDiv[i]=i;
        	for (j = i * i; j <= MAX; j += i)
            {
				is_prime[j] = false;
				MinDiv[j]=i;
    		}
    	}
	}
	for(int i=2;i<=MAX;i++) if(is_prime[i]) MinDiv[i]=i;
}
//////////////////Sieve////////////////////

int main()
{
	int x,y;
	cin>>x>>y;
	if(y-2*x>=0&&(y-2*x)%2==0&&x-(y-2*x)/2>0)
	{
		cout<<"Yes"<<endl;
	}
	else cout<<"No"<<endl;
}