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 <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <cassert>
#include <queue>
#include <map>
#include <set>
using namespace std;
using Int = int64_t;
const Int INF = 1<<30;
const Int MOD = (Int)1e9 + 7;
const Int MAX_N = (Int)1e5 + 5;
#define debug(x) cout << #x << ": " << x << endl
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2>& p) { os << p.first << " " << p.second; return os; }
template<typename T>
ostream &operator<<(ostream &os, const vector<T> &v) { for(int i = 0; i < (int) v.size(); i++) os << v[i] << (i + 1 != v.size() ? " " : ""); return os; }

void solve()
{
    int x, y; cin >> x >> y;
    for(int i = 0; i <= 100; i++)
    {
        for(int j = 0; j <= 100; j++)
        {
            int cnt = 2 * i + 4 * j;
            if(cnt == y && i + j == x)
            {
                cout << "Yes" << endl;
                return;
            }
        }
    }
    cout << "No" << endl;
}

signed main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}