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;
template <class A, class B> inline bool chmax(A &a, const B &b) { return b > a && (a = b, true); }
template <class A, class B> inline bool chmin(A &a, const B &b) { return b < a && (a = b, true); }
typedef long long ll;
typedef vector<int> vint;
typedef pair<int, int> pint;
typedef vector<long long> vlong;
typedef vector<string> vstring;
#define _GLIBCXX_DEBUG
#define vpush(a,x) a.push_back(x);
#define rep(i, n) REP(i, 0, n)
#define all(v) v.begin(), v.end()
#define REP(i, x, n) for(int i = x; i < n; i++)
const int INF = 1 << 30;
const int dx[] = {1,0,-1,0,1,1,-1,-1};
const int dy[] = {0,-1,0,1,1,-1,-1,1};
#define stp(x)  setprecision(x)
int main(){
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int x,y;
  cin>>x>>y;
  int count=0;
  int count2=0;
  if(y%2!=0){
    cout<<"No"<<'\n';
    return(0);
  }
  if(x*2>y){
    cout<<"No"<<'\n';
    return(0);
  }
  while(y>=4){
    y-=4;
    count++;
  }
  while(y>=2){
    y-=2;
    count2++;
  }
  if(count+count2<=x){
    cout<<"Yes"<<'\n';
  }
  else{
    cout<<"No"<<'\n';
  }
  return(0);
}