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)

//#pragma GCC optimize("03,unroll-loops")
//#pragma GCC target("avx,avx2,fma")

#include<bits/stdc++.h>
typedef long long ll;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ordered_set tree<int, null_type,less_equal<int>, rb_tree_tag,tree_order_statistics_node_update>
#define pb           push_back
#define ff           first
#define ss           second
#define gcd(a, b)    __gcd(a, b)
#define lcm(a, b)    ((a)*((b)/gcd(a,b)))
#define all(v)       v.begin(),v.end()
#define lllim        2147483648
#define Pi           2*acos(0.0)
#define sci(n)	     scanf("%d",&n)
#define scii(n,m)    scanf("%d%d",&n,&m)
#define scl(n)       scanf("%lld",&n)
#define scll(n,m)    scanf("%lld%lld",&n,&m)
#define pii          pair<int,int>
#define pll 	     pair<ll,ll>
#define mem(a,b)     memset(a, b, sizeof(a))
#define fill_(a,b)   fill(a,a+n,b);
#define MOD          1000000007
#define fast_cin     ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define filein       freopen("input.txt","r", stdin)
#define watch(x)     cerr << __LINE__ << ": " << #x << " = " << (x) << '\n'
#define case		 int t,cas=1;cin>>t;while(t--)
#define forn(i,a,n)  for(int i=a;a==0?i<n:i<=n;i++)



/*------------------------------Graph Moves----------------------------*/
//const int fx[]= {+1,-1,+0,+0};
//const int fy[]= {+0,+0,+1,-1};
//const int fx[]= {+0,+0,+1,-1,-1,+1,-1,+1};  // Kings Move
//const int fy[]= {-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*---------------------------------------------------------------------*/

using namespace std;
using namespace __gnu_pbds;

const int maxn=1e5+5;


void solve()
{
	ll x,y;
	cin>>x>>y;
	//ll p=(4*x-y)%2;
	
	bool f=0;
	
	for(ll i=1;i<=x;i++)
	{
		ll l=2*i +(x-i)*4;
		if(l==y)
		{
			f=1;
			break;
		}
	}
	
	if(f)
	{
		cout<<"Yes"<<endl;
		
	}
	else
	cout<<"No"<<endl;
   


}


int main()
{
    fast_cin;
    //case
    {
        //cout<<""<<cas++<<""<<endl;
        solve();
    }

    return 0;
}