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”.
Examples
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: .
题目大概意思：Drazil想去Varda家里玩，Drazil的坐标是（0，0）,Varda的坐标是（a，b）；从位置(x，y)移动一步只能到位置(x+1，y)，(x-1，y)，(x，y+1)或(x，y-1)；如果Drazil犯了一个错误，并且不可能采取正确的步骤，从他的家到Varda的家，输出NO，否则输出YES
我的理解：走完所以的步数只要有机会能到Varda家就输出YES，否则输出NO，走到Varda家至少需要a+b步，只要走的步数等于a+b，或者比a+b多2n步则可以到Varda家，输出YES
代码如下：
#include <iostream>
#include<cmath>
using namespace std;
int main()
{
int a, b, s;
cin >> a >> b >> s;
a=abs(a); b=abs(b);
if (s >= (a + b))
{
if (((s - a - b) % 2) == 0)cout << "YES" << endl;
else cout << "NO" << endl;
}
else cout << "NO" << endl;
}