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>

using namespace std;
typedef long long ll;
#define all(x) (x).begin(), (x).end()
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vector<bool>> vvb;
typedef vector<vector<int>> vvi;
typedef vector<vector<ll>> vvl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
void debugF(ostream &os) {os << endl;}
template <typename Head, typename... Tail>
void debugF(ostream &os, Head H, Tail... T) { os << " " << H; debugF(os, T...); }
#define debug(...) cout << "[" << #__VA_ARGS__ << "]:", debugF(cout, __VA_ARGS__)
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); os << (i>0 ? " ":"") << v[i++]); return os;}


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, t, i;
    cin >> n >> t;
    //x + y = n
    //2x + 4y = t
    for(i=0;i<=n;i++){
        int j = n-i;
        if((i + j == n) && (2*i + 4*j == t)) break;
    }

    if(i <= n){
        cout << "Yes\n";
    }else{
        cout << "No\n";
    }

    return 0;
}