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: .

題目大概意思：Drazil想去Varda家裡玩，Drazil的座標是（0，0）,Varda的座標是（a，b）；從位置(x，y)移動一步只能到位置(x+1，y)，(x-1，y)，(x，y+1)或(x，y-1)；如果Drazil犯了一個錯誤，並且不可能採取正確的步驟，從他的家到Varda的家，輸出NO，否則輸出YES

我的理解：走完所以的步數只要有機會能到Varda家就輸出YES，否則輸出NO，走到Varda家至少需要a+b步，只要走的步數等於a+b，或者比a+b多2n步則可以到Varda家，輸出YES
程式碼如下：

#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;
}