A. Drazil and Date

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).

Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.

Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?

Input

You are given three integers a, b, and s ( - 10⁹ ≤ a, b ≤ 10⁹, 1 ≤ s ≤ 2·10⁹) in a single line.

Output

If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).

Otherwise, print "Yes".

Sample test(s)

Input

5 5 11

Output

No

Input

10 15 25

Output

Yes

Input

0 5 1

Output

No

Input

0 0 2

Output

Yes

Note

In fourth sample case one possible route is: .

题意：给出坐标和步数 判断在使用完所有步数时能否到达指定

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
using namespace std;

int main()
{
    int x,y,s;
    cin>>x>>y>>s;
    if(abs(x)+abs(y)<=s&&(int)(s-abs(x)-abs(y))%2==0) cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
}

坐标点

分析：上下左右每次只行走一个单位 那么到达指定坐标（x，y）的最小距离一定为|x|+|y| 此时如果步数s小于则一定到不了，如果大于并且多余的步数是2的倍数则一定能准确到达；