Codeforces Round #461 (Div. 2) A. Cloning Toys

A. Cloning Toys

time limit per test 1 second
memory limit per test 256 megabytes

Problem Description

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.

Input

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).

Output

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

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

Examples

Input
6 3
Output
Yes

Input
4 2
Output
No

Input
1000 1001
Output
Yes

Note

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

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解题心得：

1. 比赛的时候差点被这个题的题意给绕进去，看了好久，其实这个题的意思是，
   
   • 你将一个原版的物品放入copy机，可以多得到一个原版的物品，多得到一个复制的物品。
   • 如果把复制的物品放入copy机，可以多得到两个复制的物品。
2. 判定是否可以出现x个复制物品，y个原版的物品就很简单了。x-y+1必须是一个偶数，x-y+1必须大于等于零，y必须大于1(不能丢弃），y等于1的时候x必须等于0。判断不完全可能被hack。

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#include <bits/stdc++.h>
using namespace std;
const int maxn = 10;
int main(){
    int a,b;
    scanf("%d%d",&a,&b);
    if((b == 1 && a != 0) || b == 0){
        printf("No
");
        return 0;
    }
    int c = a-b+1;
    if(c%2 || c < 0)
        printf("No
");
    else
        printf("Yes
");
    return 0;
}