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 Ruby (2.3.3) (judge verdict: Accepted)

n = gets.to_i

# 前に存在していた点
lx = ly = 0
# 前に存在していた点にいた時刻
lt = 0
# 合否
is_enable = true

n.times do
  t, x, y = gets.split.map(&:to_i)
  if !is_enable
    next  # 既にNGだとわかっているので読み飛ばすだけ ※最後まで読み込まないとAtCoderではRuntimeErrorになる
  end

  # 前に存在していた点からの距離（道のり）
  distance = (x - lx).abs + (y - ly).abs

  # 前に存在していた点にいた時刻からの経過時刻
  dt = t - lt

  if dt < distance
    is_enable = false
    next
  end

  # 経過時刻が偶数なら距離も偶数じゃなければならない。奇数でも同様。
  if dt.even? != distance.even?
    is_enable = false
    next
  end

  lt = t
  lx = x
  ly = y
end

puts (is_enable ? 'Yes' : 'No')