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 <algorithm>
#include <string>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <iomanip>
#include <set>
#include <ctime>
#include <tuple>
#include <bitset>
#include <assert.h>
#include <deque>
using namespace std;

/*
using PAIR = pair<string, int>;
sort(sp.begin(), sp.end(), [](PAIR l, PAIR r){
        return l.first<r.first || (l.first==r.first && l.second > r.second);
        });
*/

typedef long long ll;

#define fi first
#define se second
#define rep(i, n) for (ll i = 0; i < n; i++)
#define debugA() cerr << "AAAAA" << endl
#define debug_() cerr << "-------------" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define debug_vec(v)                   \
    cout << #v << endl;                \
    for (int i = 0; i < v.size(); i++) \
    {                                  \
        cout << v[i] << " ";           \
    }                                  \
    cout << endl;
#define debug_vec2(v)                         \
    cout << #v << endl;                       \
    for (int i = 0; i < v.size(); i++)        \
    {                                         \
        for (int j = 0; j < v[i].size(); j++) \
        {                                     \
            cout << v[i][j] << " ";           \
        }                                     \
        cout << endl;                         \
    }

using Graph = vector<vector<int>>;
using P = pair<int, int>;
using P1 = pair<int, pair<int, int>>; // クラスカル法とかで、{cost, {from, to}}的に使う。
const int INF = 1001001001;
const double pi = acos(-1);

int main()
{
    int x, y;
    cin >> x >> y;
    if (y % 2 != 0)
    {
        cout << "No" << endl;
        return 0;
    }
    if (2 * x <= y && y <= 4 * x)
    {
        cout << "Yes" << endl;
    }
    else
    {
        cout << "No" << endl;
    }

    return 0;
}