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.*;
public class Main {
	public static void main(String[] args) {
		boolean result = true;
		//出発地点の初期化
		int t0 = 0, x0 = 0, y0 = 0;
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		for (int i=0;i<n;i++) {
			//目的地の更新
			int t1 = sc.nextInt();
			int x1 = sc.nextInt();
			int y1 = sc.nextInt();
			
			int t = t1-t0;
			int d = distance(x0,x1,y0,y1);
			//時間内に届くかを判定
			if (t < d) {
				result = false;
				break;
			}
			//偶奇が一致するか判定
			if (((t^d)&1)==1) {
				result = false;
				break;
			}
			//出発地点の更新
			t0 = t1;
			x0 = x1;
			y0 = y1;
		}
		System.out.println(result ? "Yes" : "No");
	}
	
	//2地点のマンハッタン距離を返す
	public static int distance(int x1, int x2, int y1, int y2) {
		x1 -= x2;
		y1 -= y2;
		if (x1<0)
			x1 *= -1;
		if (y1<0)
			y1 *= -1;
		return x1+y1;
	}
}