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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		no1();
	}

	static void no1() {
		try (Scanner sc = new Scanner(System.in)) {
			int n = sc.nextInt();

			int[] t = new int[n];
			int[] a = new int[n];
			int[] b = new int[n];

			for (int i = 0; i < n; i++) {
				t[i] = sc.nextInt();
				a[i] = sc.nextInt();
				b[i] = sc.nextInt();
			}

			int tt = 0;
			int aa = 0;
			int bb = 0;

			int dt;
			int dd;

			for (int i = 0; i < n; i++) {
				dt = t[i] - tt;
				dd = Math.abs(a[i] - aa) + Math.abs(b[i] - bb);

				if (dd > dt) {
					System.out.println("No");
					return;
				}

				if((dd-dt) % 2 != 0) {
					System.out.println("No");
					return;
				}

				tt = t[i];
				aa = a[i];
				bb = b[i];
			}

			System.out.println("Yes");

		}
	}

}