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 Scala (2.11.7) (judge verdict: Time Limit Exceeded)

object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val n = sc.nextInt
  val list = List.fill(n, 3)(sc.nextInt)
    
  def check(term: Int, start: (Int, Int), end: (Int, Int)): Boolean = {
    val diffX = math.abs(end._1 - start._1)
    val diffY = math.abs(end._2 - start._2)
    val diff = diffX + diffY
    
    (diff <= term) && (diff % 2 == term % 2)
  }
  
  val results: Seq[Boolean] = for (index <- (0 to list.length - 1)) yield {
    if (index == 0) check(list(index)(0), (0, 0), (list(index)(1), list(index)(2)))
    else check(list(index)(0) - list(index - 1)(0), (list(index - 1)(1), list(index - 1)(2)), (list(index)(1), list(index)(2)))
  }
  
  println(if (results.contains(false)) "No" else "Yes")
}