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)

#include <bits/stdc++.h>
using namespace std;
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp> 
// using namespace __gnu_pbds;
#define ordered_set tree<pair<int,int>, null_type,less<pair<int,int> >, rb_tree_tag,tree_order_statistics_node_update>
#define standardVar ll n,m,i,j,k,num,num1,num2,flag=0
#define nn <<"\n"
#define __ <<" "<<
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define ittr(x,v) for(auto x=v.begin();x!=v.end();x++)
#define itr(x,n) for(x=0;x<n;x++)
#define itrab(x,a,b) for(x=a;x<b;x++)
#define itpr(v) ittr(x,v)cout<<*x<<" "; cout nn
#define itppr(v) ittr(x,v)cout<<(*x).ff<<" "<<(*x).ss<<"  "; cout nn
#define size(x) (ll)x.size()
#define pi 3.14159265359
#define dTor(x) ((x)*pi)/180.0
#define INF 1000000100;

typedef long long int ll;       //TYPEDEFS
typedef long double ld;
typedef vector<ll> vl;
typedef map<ll,ll> ml;
typedef set<ll> sl;
typedef string str;
typedef pair<ll, ll> pl;
typedef vector< pl > vpl;
typedef map<pl,ll> mpl;
typedef set< pl > spl;
typedef vector<str> vst;
typedef vector<vl> vvl;

const ll M=1000000007;          //COMMON MODULO PRIMES
const ll N=998244353;

ll power(ll x, ll y, ll p){     //MODULAR EXPONENTIATION
    ll res=1;
    x=x%p;
    if(x==0)
        return 0;
    while(y>0){
        if(y&1)
            res=(res*x)%p;
        y=y>>1;
        x=(x*x)%p;
    }
    return res;
}
ll inv(ll x, ll m){             //FERMAT INVERSE
    return power(x,m-2,m);
}
struct comp{                    //COMPARISON ON PAIRS
    template<typename T>        
    bool operator()(const T &a, const T &b){
        if (a.ff==b.ff)
            return (a.ss<b.ss);
        else
            return (a.ff<b.ff);
    }
};
bool mod(const pair<ll,ll> &a, const pair<ll,ll> &b){   //COMPARATOR FOR PAIRS (SORTING)
    if(a.ff!=b.ff)
        return (a.ff>b.ff);
    else
        return (a.ss>b.ss);
}
vector<bool> pr(1000000+1,1);
void sieve(ll mn, ll mx){       //SIEVE OF ERATOSTHENES
    for(ll i=mn;i*i<=mx;i++)
        if(pr[i-mn]==1)
            for(ll j=i*i;j<=mx;j+=i)
                pr[j-mn]=0;
}
bool isValid(ll i, ll j, ll n, ll m){   //CHECKS FOR OVERFLOW
    if(i>=0&&i<n&&j>=0&&j<m){
        return true;
    }
    return false;
}


//MAIN FUNCTION STARTS HERE

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    standardVar;
    cin>>n>>m;
    ll x=n-(m-2*n)/2; ll y=(m-2*n)/2;
    if(x+y==n&&x>=0&&y>=0)
        cout<<"Yes" nn;
    else
        cout<<"No" nn;
    return 0;
}