Imp likes his plush toy a lot.

Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, he additionally gets one more original toy and one copy, and if he applies the machine to a copied toy, he gets two additional copies.

Initially, Imp has only one original toy. He wants to know if it is possible to use machine to get exactly x copied toys and y original toys? He can't throw toys away, and he can't apply the machine to a copy if he doesn't currently have any copies.

The only line contains two integers x and y (0 ≤ x, y ≤ 109) — the number of copies and the number of original toys Imp wants to get (including the initial one).

Print "Yes", if the desired configuration is possible, and "No" otherwise.

You can print each letter in arbitrary case (upper or lower).

In the first example, Imp has to apply the machine twice to original toys and then twice to copies.

有两种玩具，这里假设成A、B两种玩具，A玩具可以再产生一个A玩具和一个B玩具，B玩具可以再产生两个B玩具。现在我们有一个A玩具，问最后是否能得到X个B玩具，Y个A玩具。能得到输出Yes，不能得到输出No。

#include<map>

#include<queue>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

#define maxn 100010

#define debug(a) cout << #a << " " << a << endl

using namespace std;

typedef long long ll;

int main() {

    int n,m;

    while( cin >> n >> m ) {

        if( n ==  && m ==  ) {

            cout << "Yes" << endl;

            continue;

        }

        if( n ==  || m ==  || m ==  ) {

            cout << "No" << endl;

            continue;

        }

        if( ( n - m +  ) %  ==  && ( n - m +  ) >=  ) {

            cout << "Yes" << endl;   //注意负的偶数余2也等于0，这里要特判下，因为没有考虑到这种情况WA两次。。。

        } else {

            cout << "No" << endl;

        }

    }

    return ;

}