Problem:
Score : 300 points

Problem Statement

AtCoDeer the deer is going on a trip in a two-dimensional plane.
In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i.

If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1).
Note that he cannot stay at his place.
Determine whether he can carry out his plan.

Constraints

1 ≤ N ≤ 10^5

0 ≤ x_i ≤ 10^5

0 ≤ y_i ≤ 10^5

1 ≤ t_i ≤ 10^5

t_i < t_{i+1} (1 ≤ i ≤ N-1)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
t_1 x_1 y_1
t_2 x_2 y_2
:
t_N x_N y_N

Output

If AtCoDeer can carry out his plan, print Yes; if he cannot, print No.

Sample Input 1

2
3 1 2
6 1 1

Sample Output 1

Yes

For example, he can travel as follows: (0,0), (0,1), (1,1), (1,2), (1,1), (1,0), then (1,1).

Sample Input 2

1
2 100 100

Sample Output 2

No

It is impossible to be at (100,100) two seconds after being at (0,0).

Sample Input 3

2
5 1 1
100 1 1

Sample Output 3

No

Solution in Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)



import java.util.Scanner;

public class Main{

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n, t = 0, x = 0, y = 0;
		boolean f=false;
		n = sc.nextInt();
		for (int i = 0, tt, tx, ty; i < n; i++) {
			tt = sc.nextInt();
			tx = sc.nextInt();
			ty = sc.nextInt();
			if ((Math.abs(tx - x) + Math.abs(ty - y)) % 2 != (tt - t) % 2// 判断奇偶性是否一样
					|| Math.abs(tx - x) + Math.abs(ty - y) > (tt - t)) {// 是否大于最短步数
				
				f=true;
			}
				t = tt;
				x = tx;
				y = ty;
			
		}
		if(f){
			System.out.println("No");
		}else
			System.out.println("Yes");

	}

}