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++ (Clang 10.0.0) (judge verdict: Accepted)

// #include <bits/stdc++.h>

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <string>
#include <functional>
#include <math.h>
#include <map>
#include <queue>
#include <stack>

typedef long long ll;
typedef unsigned long long ull;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define RFOR(i,a,b) for(int i=(b-1);i>=(a);--i)
#define REP(i,n)   FOR(i,0,n)
#define RREP(i,n)   RFOR(i,0,n)
#define SORT(s) sort(s.begin(), s.end())
#define P pair<int, int>
#define mp make_pair

const double INF = 1e+09;
const int MOD = 1e9+7;

using namespace std;

template <typename _Ty>
ostream& operator << (ostream& ostr, const vector<_Ty>& v) {if (v.empty()) {ostr << "{ }"; return ostr;}
    ostr << "{" << v.front(); for (auto itr = ++v.begin(); itr != v.end(); itr++) {ostr << ", " << *itr;}
    ostr << "}";
    return ostr;}

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int gcd(int a, int b){
   if (a%b == 0) return(b);
   else return(gcd(b, a%b));}

int lcm(int a, int b)
{return a * b / gcd(a, b);}

vector<int> uniq(vector<int> vec){
  sort(vec.begin(), vec.end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
  return vec;}

vector<ll> divisor(ll n) {
vector<ll> ret;
for(int64_t i = 1; i * i <= n; i++) {
  if(n % i == 0) {
    ret.push_back(i);
    if(i * i != n) ret.push_back(n / i);
  }
}
  sort(begin(ret), end(ret));
  return (ret);
}
void permu(int n){
  vector<int> perm(n);
  REP(i, n) perm[i] = i;
do{
  cout << perm  << endl;
}
while (next_permutation(perm.begin(), perm.end()));
return;}

vector<pair<ll, int>> prime_div(ll n){
  vector<pair<ll, int>> res;
  for (ll i=2; i*i <= n;i++){
    if (n%i == 0){
      int a = 0;
      while(n%i == 0){
        n /= i;
        a++;
      }
      pair<ll, int> tmp;
      tmp.first = i;
      tmp.second = a;
      res.push_back(tmp);
    }
  }
  if (n != 1){
    pair<ll, int> now;
    now.first = n;
    now.second = 1;
    res.push_back(now);}
  return res;
}


// printf("%.10f\n" , ans);
int main(void){
  ios::sync_with_stdio(false);
  int x, y;
  cin >> x >> y;
  REP(i, x+1){
    int val = i * 2 + (x-i) * 4;
    if (val == y){
      cout << "Yes" << endl;
      return 0;
    }
  }

cout << "No" << endl;
  return 0;
}