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 Python (3.8.2) (judge verdict: Runtime Error)

#13 C - Traveling
N = int(input())

plan = []
for _ in range(N):
    ti,xi,yi = map(int,input().split())
    plan.append((ti,xi,yi))
    
#今いる場所の座標と時間
spot = (0,0,0)#始点
result = 'Yes'
for t,x,y in plan:
    tdiff = t-spot[0]
    xdiff = abs(x-spot[1])
    ydiff = abs(y-spot[2])
    #(x,y)の変化量の和の偶奇と時間の差の偶奇が一致している
    #時間の変化量より(x,y)の変化量の方が小さい
    if (tdiff%2 == (xdiff+ydiff)%2) and ((xdiff + ydiff)<=tdiff):
    '''
    xdiff<=tdiff and ydiff<=t だと(3,3,2)などの範囲外に対応できない
    if (tdiff%2 == (xdiff+ydiff)%2) and (xdiff<=tdiff) and (ydiff<=tdiff):
    '''
        spot = (t,x,y)
    else:
        result = 'No'
        break
print(result)