Codeforces Round #292 (Div. 2) -- A. Drazil and Date

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: 

.

注意：a，b這裡可以取負數，是以要加個abs

AC代碼：

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

int main() {
	int a, b, s;
	while(scanf("%d %d %d", &a, &b, &s) != EOF) {
		int t = abs(a) + abs(b);
		if(t == 0 && !(s & 1))
			printf("Yes\n");
		else if(t == 0 && s & 1) 
			printf("No\n");
		else if(t > s) printf("No\n");
		else if( (s - t) & 1) printf("No\n");
		else printf("Yes\n");
	}
	return 0;
}