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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MOD1 1000000007
#define MOD2 998244353
#define LIMIT1 200002
#define LIMIT2 500002
typedef long long ll;
typedef long double ld;
typedef const void cv;
#define rep(i,n) for((i)=0;(i)<(n);(i)++)
#define per(i,n) for((i)=(n)-1;(i)>=0;(i)--)
#define max(a,b) ((a)>(b) ? (a) : (b))
#define min(a,b) ((a)<(b) ? (a) : (b))
#define zt(a,b) (max((a),(b))-min((a),(b)))
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll nPr(int n,int r){ll i,result=1;rep(i,r){result*=(ll)(n-i);}return result;}
ll nCr(int n,int r){ll i,result=1;if(n<r) return 0;rep(i,min(r,n-r)){result*=(ll)(n-i);result/=(ll)(i+1);}return result;}
#define nHr(n,r) nCr((int)((n)+(r)+1),(int)(r))
#define sankaku(x) (((x)*((x)+1))/2)
int dx[8]={1,0,-1,0,1,-1,-1,1},dy[8]={0,1,0,-1,1,1,-1,-1};
int upint(cv *a, cv *b) { return *(int *)a < *(int *)b ? -1 : *(int *)a > *(int *)b ? 1 : 0; }
int downint(cv *a, cv *b) { return *(int *)a < *(int *)b ? 1 : *(int *)a > *(int *)b ? -1 : 0; }
int upchar(cv *left, cv *right) {return strcmp((char *)left,(char *)right);}
int downchar(cv *left, cv *right) {return strcmp((char *)right,(char *)left);}
void tmpInit(int *c,ll m) {ll i;rep(i,m) c[i]=i;}
ll modpow(ll a, ll n, ll mod) {
  ll res = 1;
  while (n > 0) {
    if (n & 1){
      res *= a; if(mod>=1) res %= mod;
    }
    a *= a; if(mod>=1) a %= mod;
    n >>= 1;
  }
  return res;
}
void initialize(){
  
}
int main(void){
  initialize();
  ll n,m,i,j,k,result=0;
  int a[LIMIT2]={0};
  char s[LIMIT1];
  
  scanf("%lld %lld",&n,&m);
  if(2*n>=m && 4*n<=m && m%2==0) printf("Yes\n");
  else printf("No\n");
  
  return 0;
}